Olympiad Test Level 2: Logical Reasoning - Class 6 MCQ

Let the smaller number be x.
Then larger number = (x + 1365).
∴ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270
Smaller number = 270.

Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.
∴ x = 5k + 3
∴ x² = (5k +3)²
= (25k² + 30k + 9)
= 5(5k² + 6k+1) + 4
∴ on dividing x² by 5, we get 4 as remainder.

Final position of Lokesh is at point ‘E’ which is north from his house.

Required average runs scored by MS Dhoni

Required average runs scored by S Raina 

Required average runs scored by R Sharma 

Required average runs scored by H Pandya 

The required cricketer is M.S. Dhoni.

Required total score = 60 + 50 + 55 + 50 = 215 

1, 5, 7 contain a rectangle with its two diagonals as the outer element and another element (similar or different) placed inside it.
2, 6, 9 contain a triangle with its three medians as the outer element and another element (similar or different) placed inside it. 
3, 4, 8 contain a circle with its two mutually perpendicular diameters as the outer element and another element (similar or different) placed inside it.

L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3 ) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.

Quantity of M ilk = 60 x (2/3) = 40 liters Quantity of water = 60 - 40 = 20 liters
As per question we need to add water to get quantity 1:2
⇒ 40/(20 + x) = 1/2
⇒ 20 + x = 80
⇒ x = 60 liters

According to question,
Area of road that is parallel to length
= 120 x 15 = 1800 sq m
Area of road that is parallel to breadth
= 90 x 15 = 1350 sqm
Area of intersection of two roads
= 15 x 15 = 22m²
∴ Area of two road = (1800+ 1350-225)
= 2925 sq m
Cost at 1 sq m = 20
So, Cost at 2925 sq m = 20 x 2925
= ₹ 58500

According to question

so d = 13

Are of  path = 2 x (1 + b - 2x)
= 5 (112 + 78 - 5)
= 5 * 185 = 925 sq m
Therefore, Required cost = (925x2)
= ₹ 1850

Step 1: Findarea of inner square
= 2 cm x 2 cm = 4 cm²
Step 2 : Find area of outer shape
= (2 cm x 3 cm) + (10 cm x 3 cm)
= 6 cm² + 30 cm² = 36 cm²
Step 3 : Area of shaded region
= area of outer shape - area of inner square
= 36 cm² - 4 cm² = 32 cm²

Here always remember, when ever, x% loss,
it means S.P. = (100 - x) % of C.P
When ever, x% profit, it means S.P. = (100 + x) % of C.P
So here will be (100 - x) % of C.P.
= 80 % of 1200
= 80/100 x 1200
= 960

Required distance