Please wait...

Ratio & Proportion Test 158
Result
Ratio & Proportion Test 158

  • /

    Score

  • -

    Rank
Time Taken: -
  • Question 1/10
    1 / -0.25
    Find two numbers such that their mean proportion is 12 and third proportion is 768.
    Solutions
    Let the numbers are x and y, then

    ⇒ x = 144/y ---------------(i)
    Given: third proportion = 768, then

    (By eq. (i))
    ⇒ y3 = 110592
    ⇒ y = 48
    Putting this value in eq. (i), we get
    x = 3
    Thus, required numbers are 3 and 48.
  • Question 2/10
    1 / -0.25
    If the ratio of Income of A, B and C is 5:7:9 and their saving ratio is 3:4:5 and ratio of A’s income to A’s savings is 5:9. Find the ratio of total income of A, B & C’s income to total savings of all three.
    Solutions




    Put
    Now,

    Required,

  • Question 3/10
    1 / -0.25
    In an MBA selection process, the ratio of selected to unselected was 11:2. If 40 less had applied and 20 less selected, the ratio of selected to unselected would have been 10:1. How many candidates had applied for the process?
    Solutions
    Let the selected candidates = 11x and Unselected candidates = 2x

    Total candidates = 11x + 2x = 13x

    New number of candidates who applied = 13x - 40

    New number of candidates who were selected = 11x - 20

    Ratio of the selected candidates to unselected candidates = 10:1

    Ratio of total candidates to selected candidates = 11:10

    So, 

    ⇒ 130x - 400 = 121x - 220

    ⇒ 9x = 180     ⇒ x = 20

    Hence, Total number of candidates who applied = 13x = 13 × 20 = 260
  • Question 4/10
    1 / -0.25
    There are two buckets, smaller bucket can hold only 3/5th of the water as compared to the larger bucket. If 6,000 buckets of larger capacity are needed to fill a pond. Then how many smaller capacity buckets are needed to fill the same pond?
    Solutions
    Let the capacity of smaller and bigger buckets are c and C respectively.
    Given, c = 3/5 C
    To fill the ponds we need 6000 bugger buckets.
    So, Capacity of pond = 6000 C
    = 6000 (5/3) c
    = 10000 c
    Hence, 10000 of smaller capacity buckets are needed to fill the pond.
  • Question 5/10
    1 / -0.25
    The ratio of number of cans of orange, pineapple and mixed fruit juices kept in a store is 8 : 9 : 15. If the store sells 25%, 33.33% and 20% of orange, pineapple and mixed fruit juices cans respectively, then what is the ratio of number of cans of these juices in the remaining stock?
    Solutions
    Ratio of number of cans of orange, pinapple and mixed fruit = 8:9:15

    Let the number of orange cans = 80

    Number of pineapple = 90

    Number of mixed fruit = 150

    Remaining cans of orange = 80-25% of 80 = 75% of 80 =

    Remaining cans of pineaaple = 90-33.33% of 90 =

    Remaining cans of mixed fruit = 150-20% of 150

    = 80% of 150 =

    Required ratio = 60:60:120 = 1:1:2
  • Question 6/10
    1 / -0.25
    The cost of diamond varies directly as the square of its weight, once this diamond broke into 4 pieces with weight in ratio 1:2:3:4, If the pieces were sold, the dealer would get Rs.70,000 less. Find the original price of the diamond:
    Solutions
    Given that,
    Price ∝ Weight2
    ⇒ p ∝ (w2)
    Let the weight of the diamond = 10x
    ⇒ p = k (10x)2
    ⇒ p = k (100x2) ------ (i)
    Since the diamond has been broken into 4 pieces
    10x will get divided into the ratio of 1:2:3:4
    P1 = k(x2), p2 = k (4x2), P3 = k (9x2), P4 = k (16x2)
    P1 + p2 + p3 + p4 = k (30x2)
    According to the question,
    ⇒ k (100x2) - k (30x2) = 70000
    ⇒ k70x2 = 70000
    ⇒ kx2 = 1000
    From eq (i):
    ∴ Price = 1000 × 100 = Rs.1,00,000
  • Question 7/10
    1 / -0.25
    A bag has Rs.43 in the form of 5 rupees, 50 paise and 10 paise coins in the ratio of 1:5:11. What is the total number of 50 paise coins?
    Solutions
    Face value of 5 rupees, 50 paise, and 10 paise are in the ratio of



    Number of coins are in the ratio of 1:5:11

    Let the number of 5 rupee coins = x

    50 paise = 5x

    And 10 paise = 11x

    Then, 5x + 5x × 0.5 + 11x × 0.1 = 43

    ⇒ 5x + 2.5x + 1.1x = 43

    ⇒ 8.6x = 43

    ⇒ x = 5

    Number of 50 paise coin = 5x = 5 × 5 = 25
  • Question 8/10
    1 / -0.25
    A student obtained equal marks in Mathematics and Physics. The ratio of marks in Physics and Chemistry is 2:3 and the ratio of marks in Mathematics and English is 1:2. If he has scored an aggregate of 66% marks. The maximum marks in each subject is same. In how many subjects he scored 72% and above marks.
    Solutions
    Let the marks in maths = physics = 2x

    Marks in chemistry = 3x

    Marks in English = 4x

    Aggregate marks = 2x+2x+3x+4x / 4 = 11x/4 = 66%

    X= 24%

    Thus marks in maths = physics = 2x = 48%

    Marks in chemistry = 3x = 72%

    Marks in English = 4x = 96%

    Thus in English and chemistry he scored 72% or more

  • Question 9/10
    1 / -0.25
    In a city 20% of male population is 35% of female population 40% of males are married and 70% of them have children. 60% of females are married and 75% of them have children then what is the respective ratio of males to females who haven’t a children?
    Solutions
    No of males: No of females = 7:4
    Required ratio =
  • Question 10/10
    1 / -0.25
    Rs. 5040 is divided among 5 men, 4 women and 3 boys. The ratio of share of a man, a woman and a boy is 7:4:3. What is the share of a woman in rupees?
    Solutions
    let the share be 7x, 4x and 3x respectively

    Now 7x × 5 + 4x × 4 + 3x × 3 = 5040

    60x = 5040

    Or x = 84

    Share of a woman = 4x = 336

User Profile
-

Correct (-)

Wrong (-)

Skipped (-)

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Click on Allow to receive notifications
×
Open Now