High Order Thinking Skills -1 for Class 7 Mathematics Olympiad - Free MCQ

LCM of 16, 36, 80 = 720 Required smallest number = 720 + 7 = 727

2 + 12 < 15, 
So, 2, 12, 15 cannot be sides of a triangle.

Given:
Raju runs 3 rounds
Total distance = 7 km 800 m = 7,800 m
Width of the stadium = 330 m
Let length = L

Step 1: Find the perimeter
Total distance in 3 rounds = 3 × (Perimeter)
7800 = 3 × Perimeter
Perimeter = 7800 ÷ 3 = 2600 m

Step 2: Use perimeter formula of a rectangle
Perimeter = 2 × (Length + Width)
2600 = 2 × (L + 330)
2600/2 = L + 330
1300 = L + 330
L = 1300 - 330
L = 970 m

Correct Option: D

Length of rectangular floor of room= 9m

breadth of rectangular floor= 6m

∴Area of rectangular floor = length* breadth

                                     =   9*6

                                       = 54 m²

Side of Square tile= 15cm

Now we need to convert Side of Square tile in meters (since unit of all quantity has to be same for ease of calculation)= 15/100=0.15m

Area of square tile = side²

                             = 0.15²

                            =0.0225 m²

Now number of tiles required to cover the floor =(area of floor of room)/(area of 1 tile)

 = (54)/(0.0225)

=2400 tiles

2x + 3x + 5x = 7140
⇒ 10x = 7140
⇒ x = 7140 ÷ 10 = 714
So, Share of B = 714 × 3 = 2142

Let the breadth be b cm
Length = (b + 7) cm
Perimeter = 2 (l + b)
= 2 (b + 7 + b) = 2 (2b + 7)
⇒ 2 (2b + 7) = 172
⇒ 2b + 7 = 86

Let three consecutive even numbers be x, x + 2, x + 4. 
∴ x + x + 2 + x + 4 = 222
⇒ 3x + 6 = 222
⇒ 3x = 222 – 6
⇒ 3x = 216

Largest even number = x + 4
= 72 + 4 = 76

∵ In 12 days a hole can be dig by 26 men 
∴ In 1 day a hole can be dig by 26 × 12 men
∴ In 4 days a hole can be dig by =  men
= 78 men 

∵ In 11 litres a car covers 121 km. 
∵ In 1 litre a car covers 121/11 = 11 km.
∴ In 15 litres a car covers 11 × 15 = 165 km.

∴ The cost of 5 bars of soap = Rs. 92.50
∴ The cost of 1 bar of soap = 
∴ The cost of 36 bars of soap = 

= 666

x + 3x + 5x = 90

∴ Length of smallest side = x = 10cm 

∵ 550 men had provision for 56 days. 
∴ 1 man had provision for 56 × 550
∴ 700 men had provision for 
= 44 days 

Perimeter of square = 

Area = (32)² = 1024 m² 

Distance covered by walking  18km – [8 km 65 m + 4 km 325 m] 18km – 12km 390m = 5km 610m

[14-{12-{9-(7-4) } ] 

[14- {12-{9- (3) } ] 

[14- {12- {9-3 } ] 

[14- {12- { 6 } ] 

[14- {12- 6 }] 

[14 - { 6 }] 

= 8

LCM of 12 and 18 = 36 minutes. 6 a.m. & 36 minutes = 6 : 36 A.M.

∴ 3 × 3 × 4 × 7 × 5
= 420 × 3 = 1260
LCM of 3, 7, 9, 12, 15, 18 = 1260 

2 (l + b) = 46 ⇒l + b = 23 ……..(1) 
lb = 120

From (1) & (2) 2l = 30 ⇒l = 15 
b = 8
∴ Length of diagonal

Find the area of the sheet of paper

The sheet of paper has dimensions 95 cm by 72 cm. The area of the sheet is:

Area of sheet = 95 cm × 72 cm = 6840 cm²

Find the area of one envelope

Each envelope requires a paper of size 19 cm by 18 cm. The area of one envelope is:

Area of one envelope = 19 cm × 18 cm = 342 cm²

Calculate how many envelopes can be made

Now, divide the total area of the sheet by the area required for one envelope:

Number of envelopes = 6840342 = 20

Conclusion:

You can make 20 envelopes out of the sheet of paper.

Step 1: Solve inside the parentheses

Step 2: Add 4/3

Step 3: Add 6