44089.In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:
20 litres
30 litres
40 litres
60 litres
Explanation:
Quantity of milk = $ (60 \times \dfrac{2}{3}) litres = 40 litres.$
Quantity of water in it = (60- 40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then, milk : water = ($ \dfrac{40}{20+x}$)
now, ($ \dfrac{40}{20+x}$) = $\dfrac{1}{2}$
=> 20 + x = 80
=> x = 60.
Quantity of milk = $ (60 \times \dfrac{2}{3}) litres = 40 litres.$
Quantity of water in it = (60- 40) litres = 20 litres.
New ratio = 1 : 2
Let quantity of water to be added further be x litres.
Then, milk : water = ($ \dfrac{40}{20+x}$)
now, ($ \dfrac{40}{20+x}$) = $\dfrac{1}{2}$
=> 20 + x = 80
=> x = 60.
44090.The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
3 : 3 : 10
10 : 11 : 20
23 : 33 : 60
Cannot be determined
Explanation:
Let A = 2k, B = 3k and C = 5k
A's new salary = $\dfrac{115}{100} of 2 k = (\dfrac{115}{100}\times 2 k ) = \dfrac{23 k}{10}$
B's new salary = $\dfrac{110}{100} of 3 k = (\dfrac{110}{100}\times 3 k ) = \dfrac{33 k}{10}$
C's new salary = $\dfrac{120}{100} of 5 k = (\dfrac{120}{100}\times 5 k ) = 6 k$
ஃ New ratio ($\dfrac{23k}{10} : \dfrac{33k}{10}: 6 k) $= 23 : 33 : 60
A's new salary = $\dfrac{115}{100} of 2 k = (\dfrac{115}{100}\times 2 k ) = \dfrac{23 k}{10}$
B's new salary = $\dfrac{110}{100} of 3 k = (\dfrac{110}{100}\times 3 k ) = \dfrac{33 k}{10}$
C's new salary = $\dfrac{120}{100} of 5 k = (\dfrac{120}{100}\times 5 k ) = 6 k$
ஃ New ratio ($\dfrac{23k}{10} : \dfrac{33k}{10}: 6 k) $= 23 : 33 : 60
44091.Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
2 : 5
3 : 5
4 : 5
6 : 7
Explanation:
Let the third number be x.
Then, first number = 120% of x =$\dfrac{120x}{100}=\dfrac{6x}{5}$
Second number = 150% of x = $\dfrac{150x}{100}=\dfrac{3x}{2}$
Ratio of first two numbers = ($\dfrac{6x}{5} : \dfrac{3x}{2}$)= 12x : 15x = 4 : 5.
Let the third number be x.
Then, first number = 120% of x =$\dfrac{120x}{100}=\dfrac{6x}{5}$
Second number = 150% of x = $\dfrac{150x}{100}=\dfrac{3x}{2}$
Ratio of first two numbers = ($\dfrac{6x}{5} : \dfrac{3x}{2}$)= 12x : 15x = 4 : 5.
44092.A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B s share?
Rs. 500
Rs. 1500
Rs. 2000
None of these
Explanation:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x - 3x = 1000
x = 1000.
B s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x - 3x = 1000
x = 1000.
B s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
44093.The fourth proportional to 5, 8, 15 is:
18
24
19
20
Explanation:
Let the fourth proportional to 5, 8, 15 be x.
Then, 5 : 8 : 15 : x
=> 5x = (8 x 15)
x = $\dfrac{(8 \times15)}{5} = 24$
Let the fourth proportional to 5, 8, 15 be x.
Then, 5 : 8 : 15 : x
=> 5x = (8 x 15)
x = $\dfrac{(8 \times15)}{5} = 24$
44094.If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
2 : 5
3 : 7
5 : 3
7 : 3
Explanation:
Let 40% of A = $\dfrac{2}{3}B$
then, $\dfrac{40A}{100} = \dfrac{2B}{3}$
=> $\dfrac{2A}{5} = \dfrac{2B}{3}$
=> $\dfrac{A}{B} = (\dfrac{2}{3}\times\dfrac{5}{2}) = \dfrac{5}{3}$
ஃ A : B = 5 : 3.
Let 40% of A = $\dfrac{2}{3}B$
then, $\dfrac{40A}{100} = \dfrac{2B}{3}$
=> $\dfrac{2A}{5} = \dfrac{2B}{3}$
=> $\dfrac{A}{B} = (\dfrac{2}{3}\times\dfrac{5}{2}) = \dfrac{5}{3}$
ஃ A : B = 5 : 3.
44095.A and B together have Rs. 1210. If 4/15 of A s amount is equal to 2/5 of B s amount, how much amount does B have?
Rs. 460
Rs. 484
Rs. 550
Rs. 664
Explanation:
$\dfrac{4}{15}A = \dfrac{2}{5}B$
$=>A=(\dfrac{2}{5}\times\dfrac{15}{4})B$
$=>A=(\dfrac{3}{2})B$
$=>\dfrac{A}{B}=>\dfrac{3}{2}$
=>A:B=3:2
B s share = Rs. $(1210\times\dfrac{2}{5})=Rs.484.$
$\dfrac{4}{15}A = \dfrac{2}{5}B$
$=>A=(\dfrac{2}{5}\times\dfrac{15}{4})B$
$=>A=(\dfrac{3}{2})B$
$=>\dfrac{A}{B}=>\dfrac{3}{2}$
=>A:B=3:2
B s share = Rs. $(1210\times\dfrac{2}{5})=Rs.484.$
44096.If Rs. 782 be divided into three parts, proportional to $\dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4}$ , then the first part is:
Rs. 182
Rs. 190
Rs. 196
Rs. 204
Explanation:
Given ratio =$ \dfrac{1}{2}$ : $\dfrac{2}{3} $: $\dfrac{3}{4} $= 6 : 8 : 9.
1st part = Rs. $(782 \times \dfrac{6}{23}) $= Rs. 204
Given ratio =$ \dfrac{1}{2}$ : $\dfrac{2}{3} $: $\dfrac{3}{4} $= 6 : 8 : 9.
1st part = Rs. $(782 \times \dfrac{6}{23}) $= Rs. 204
44097.The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:
20
30
48
58
Explanation:
Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = $(5 \times\dfrac{3}{5}) : (8 \times\dfrac{3}{5}) = 3 : \dfrac{24}{5}$
=> A : B : C = 2 : 3 : $\dfrac{24}{5}$ = 10 : 15 : 24
=> B = $(98 \times \dfrac{15}{49}) = 30$
=> B = 30
Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = $(5 \times\dfrac{3}{5}) : (8 \times\dfrac{3}{5}) = 3 : \dfrac{24}{5}$
=> A : B : C = 2 : 3 : $\dfrac{24}{5}$ = 10 : 15 : 24
=> B = $(98 \times \dfrac{15}{49}) = 30$
=> B = 30
44098.Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
2 : 3 : 4
6 : 7 : 8
6 : 8 : 9
None of these
Explanation:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
=> $(\dfrac{140}{100} \times5x), (\dfrac{150}{100} \times7x) and (\dfrac{175}{100} \times8x)$
The required ratio = 7x : $\dfrac{21x}{2}$: 14x
=> 14x : 21x : 28x
=> 2 : 3 : 4.
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
=> $(\dfrac{140}{100} \times5x), (\dfrac{150}{100} \times7x) and (\dfrac{175}{100} \times8x)$
The required ratio = 7x : $\dfrac{21x}{2}$: 14x
=> 14x : 21x : 28x
=> 2 : 3 : 4.