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- Question
Two solutions have milk & water in the ratio 7:5 and 6:11. Find the proportion in which these two solutions should Be mixed so that the resulting solution has 1 part milk and 2 parts water ?
Options- A. 7 : 12
- B. 8 : 13
- C. 9 : 4
- D. 2 : 5
- Correct Answer
- 8 : 13
ExplanationTo get the solution that contains 1 part of milk and two parts of water,
they must be mixed in the ratio as
7x+6x/5y+11y = 1/2
26x = 16y
x/y = 16/26
x/y = 8/13 - 1. There are two containers, the first one contains 1 litre pure water and the second one contains 1 litre pure milk.Now 5 cups of water from the first container is taken out is mixed well in the second container. Then, 5 cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of milk in the first container and B denote the proportion of water in the second container then:
Options- A. A
- B. A=B
- C. A>B
- D. can't be determined Discuss
- 2. Maneela, Raghu and Aravind have some jems with each of them. Five times the number of jems with Raghu equals seven times the number of jems with Maneela while five times the number of jems with Maneela equals seven times the number of jems with Aravind. What is the minimum number of jems that can be there with all three of them put together ?
Options- A. 108
- B. 107
- C. 109
- D. 110 Discuss
- 3. 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 ?
Options- A. 1 : 2 : 3
- B. 2 : 3 : 4
- C. 3: 4 : 5
- D. 4 : 5 : 6 Discuss
- 4. Two vessel P and Q contain milk and water in ratio 3 : 2 and 5 : 3 respectively. If 10 liters of the mixture is removed from vessels P and poured in vessel Q then ratio of milk to water in vessel Q becomes 8 : 5. Find the initial quantity of water in vessel Q.
Options- A. 10 lit
- B. 6 lit
- C. 16 lit
- D. 12 lit Discuss
- 5. Rs. 2600 is divided into three parts A, B and C. How much A is more than C if their ratio is 1/2:1/3:1/4 ?
Options- A. 600
- B. 300
- C. 150
- D. 75 Discuss
- 6. 64 boys and 40 girls form a group for social work. During their membership drive, the same number of boys and girls joined the group. How many members does the group have now, if the ratio of boys to girls is 4:3 ?
Options- A. 168
- B. 201
- C. 147
- D. 154 Discuss
- 7. The ratio of students in a coaching preparing for B.tech and MBA is 4 : 5. The ratio of fees collected from each of B.tech and MBA students is 25 : 16. If the total amount collected from all the students is 1.62 lakh, what is the total amount collected from omly MBA aspirants?
Options- A. Rs. 62,000
- B. Rs. 72,000
- C. Rs. 82,000
- D. None of these Discuss
- 8. A man spend Rs. 810 in buying trouser at Rs. 70 each and shirt at Rs. 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased ?
Options- A. 1:3
- B. 2:1
- C. 3:2
- D. 2:3 Discuss
- 9. A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year ?
Options- A. 21:29
- B. 41:54
- C. 32:45
- D. 37:51 Discuss
- 10. In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case ?
Options- A. 693
- B. 741
- C. 528
- D. 654 Discuss
Ratio and Proportion problems
Search Results
Correct Answer: A=B
Explanation:
Here the ratio of mixtures( i.e milk , water) doesnot matter. But the important point is that whether the total amount ( either pure or mixture ) being transferred is equal or not.Since the total amount ( i.e 5 cups) being transferred from each one to another , hence A =B.
Correct Answer: 109
Explanation:
From given data,
R : M = 5 : 7
M : A = 5 : 7
R : M : A = 25 : 35 : 49
25 + 35 + 49 = 109
Correct Answer: 2 : 3 : 4
Explanation:
Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8.
Seats for Mathematics, Physics and Biology in a school are 5x, 7x and 8x.
There is a proposal to increase these seats by 40%, 50% and 75% respectively.
Then Seats for Mathematics, Physics and Biology in a school will be
1.4 X 5x, 1.5 X 7x and 1.75 X 8x and
ratio will be 7 : 10.5 : 14 => 2:3:4.
Correct Answer: 10 lit
Explanation:
Given ratio of initial mixture of milk and water in Q = 5 : 3
Let the initial quantity of mixture in vessel Q = 8x
Let quantity of Milk = 5x and
Let quantity of water = 3x
According to the question,
=> 25x + 30 = 24x + 32
=> x = 2
Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.
Correct Answer: 600
Explanation:
1/2:1/3:1/4 = 6:4:3
As the difference is 3
=> 3/13x2600 = 600
Correct Answer: 168
Explanation:
Let us say x boys and x girls joined the group.
(64 + x)/(40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168.
Correct Answer: Rs. 72,000
Explanation:
The ratio of fees collected from B.Tech : MBA = 4x * 25y : 5x * 16y
= 100xy : 80xy
= 5xy : 4 xy = 5k : 4k
The amount collected only from MBA students = = Rs. 72,000
Correct Answer: 3:2
Explanation:
Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7
We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7
Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9
Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.
Correct Answer: 32:45
Explanation:
A : B
(8000x4)+(4000x8) : (9000x6)+(6000x6)
64000 : 90000
=> 32 : 45
Correct Answer: 528
Explanation:
Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k
As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)
x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.
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