Olympiad Test: Ratio And Proportion - 3


10 Questions MCQ Test Math Olympiad for Class 5 | Olympiad Test: Ratio And Proportion - 3


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This mock test of Olympiad Test: Ratio And Proportion - 3 for Class 5 helps you for every Class 5 entrance exam. This contains 10 Multiple Choice Questions for Class 5 Olympiad Test: Ratio And Proportion - 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Olympiad Test: Ratio And Proportion - 3 quiz give you a good mix of easy questions and tough questions. Class 5 students definitely take this Olympiad Test: Ratio And Proportion - 3 exercise for a better result in the exam. You can find other Olympiad Test: Ratio And Proportion - 3 extra questions, long questions & short questions for Class 5 on EduRev as well by searching above.
QUESTION: 1

Two numbers are in ratio 4 : 5 and their LCM is 180. The smaller number is

Solution:

36 
Let two numbers are 4x and 5x;
their LCM=180 and HCF= x;
Now, 1st number × 2nd number = LCM × HCF;
⇒ 4x × 5x = 180 × x;
⇒ 20x =180;
⇒ x = 9;
then, the smaller number= 4 × 9 = 36; 

QUESTION: 2

If A : B = 2 : 3, B : C = 4 : 5 and C : D =5 : 9  then A : D is equal to

Solution:

8 : 27 
A/D= (A/B) × (B × C) × (C/D)
= (2/3) × (4/5) × (5/9)
=(2 × 4 × 5)/(3 × 5 × 9)
=8/27. 

QUESTION: 3

What must be added to each term of the ratio 7 : 11, So as to make it equal to 3 : 4?

Solution:

6.5 
Let × be added to each term.
According to question,
(7+x)/(11+x)= 3/4;
⇒ 33+3x= 28+4x;
⇒ x= 5. 

QUESTION: 4

Two numbers are in ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. The smaller number is

Solution:

49 
Let numbers are 7x and 11x.
According to question,
(7x+7)/(11x+7)=2/3;
Or, 22x+14=21x+21;
Or, x=7;
Smaller number = 7 × 7 = 49. 

QUESTION: 5

The ratio of water and milk in a 30 liter mixture is 7 : 3. Find the quantity of water to be added to the mixture in order to make this ratio 6 : 1.

Solution:

33 
Let water= 7x and milk= 3x. This will give us water in the mixture as 21 liter and 9 liter milk.
Now, we keep milk constant and add water to mixture to get ratio 6:1.
This will give 54 liter and of water and 9 liter milk.
Then extra water to be added is 33 liter. 

QUESTION: 6

The difference between two positive numbers is 10 and the ratio between them is 5 : 3. Find the product of the two numbers.

Solution:

375 
Let two positive is 5x and 3x respectively.
According to question,
5x – 3x=10; Or, x=5.
Then numbers are 25 and 15.
Thus their product= 25 × 15= 375.

QUESTION: 7

A cat leaps 5 leaps for every 4 leaps of a dog, but 3 leaps of the dog are equal to 4 leaps of the cat. What is the ratio of the speed of the cat to that of the dog?

Solution:

15 : 16 
Given :
3 dog = 4 cat;
Or, dog/cat = 4/3;
Let cat’s 1 leap = 3
meter and dogs 1 leap = 4 meter.
Then, ratio of speed of cat and dog
= 3 × 5/4 × 4= 15 : 16. 

QUESTION: 8

The present ratio of ages of A and B is 4 : 5. 18 years ago, this ratio was 11 : 16. Find the sum total of their present ages.

Solution:

90 years
Let present age of A and B is 4x and 5x.
18 years ago their ages;
(4x – 18)/ (5x – 18)=11/16;
Or, 64x – 288= 55x – 198;
Or, 64x – 55x= – 198+288;
Or, 9x=90;
Or, x= 90/9=10;
Sum of present ages= 40+50
=90 years. 

QUESTION: 9

A dishonest milk man mixed 1 liter of water for every 3 liters of milk and thus made up 36 liters of milk. If he now adds 15 liters of milk to mixture, find the ratio of milk and water in the new mixture.

Solution:

14 : 3 
Quantity of milk and water in the 36 liter mixture;
27 milk liter and 9 liter water.
Now, 15 liters milk is added then milk becomes 42 liters.
Now, ratio= 42:9=14:3 

QUESTION: 10

If the ratio of the ages of Maya and Chhaya is 6 : 5 at present, and fifteen years from now, the ratio will get changed to 9 : 8, then find Maya’s present age.

Solution:

30 years Let Maya and Chhaya’s present age is 6x and 5x respectively.
And, (6x+15)/(5x+15)=9/8;
Or, 48x+120= 45x+135;
Or, 3x=15; Or, x= 5;
∴ Present age of Maya
= 6x= 6 × 5 = 30 years.

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