Olympiad Test: Ratio And Proportion - 3 - Class 5 MCQ

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; 

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

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

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. 

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. 

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.

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. 

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. 

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 

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.