Olympiad Test : Playing With Numbers - 2 - Class 6 MCQ

# Olympiad Test : Playing With Numbers - 2 - Class 6 MCQ

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## 20 Questions MCQ Test Maths Olympiad Class 6 - Olympiad Test : Playing With Numbers - 2

Olympiad Test : Playing With Numbers - 2 for Class 6 2024 is part of Maths Olympiad Class 6 preparation. The Olympiad Test : Playing With Numbers - 2 questions and answers have been prepared according to the Class 6 exam syllabus.The Olympiad Test : Playing With Numbers - 2 MCQs are made for Class 6 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test : Playing With Numbers - 2 below.
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Olympiad Test : Playing With Numbers - 2 - Question 1

### The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725. What is the other number?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 1

= 435

Olympiad Test : Playing With Numbers - 2 - Question 2

### Which of the following is a composite number?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 2

∴ 32 has 1, 2, 4, 8, 16, 32, as its factors.
∴ 32 is a composite number.

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Olympiad Test : Playing With Numbers - 2 - Question 3

### Which longest tape can be used to measure exactly the length 7m, 3m 85cm and 12m 95 cm?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 3

Required length = HCF of 700, 385 and 1295  = 35cm

Olympiad Test : Playing With Numbers - 2 - Question 4

The greatest number that will divide 445, 572 & 699 leaving remainder 4, 5, 6 respectively.

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 4

HCF of 441, 567 and 693 = 3 × 3 × 7 = 63
∴ Required number = 63

Olympiad Test : Playing With Numbers - 2 - Question 5

What is the sum of LCM & HCF of 1152 & 1664?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 5

HCF of 1152 and 1664

∴ HCF of 1152 and 1664 = 2= 128 and LCM of 1152 and 1664 = 14976

∴ LCM + HCF = 14976 + 128 = 15104.

Olympiad Test : Playing With Numbers - 2 - Question 6

The HCF of two numbers is 21 and their LCM is 3003. If one of the numbers is 231. Then what is the other number?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 6

Required number =
= 273

Olympiad Test : Playing With Numbers - 2 - Question 7

Find the greatest 3-digit number which is divisible by 8, 10 and 12.

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 7

LCM of 8, 10 and 12

∴ LCM of 8, 10, 12 = 2 × 2 × 2 × 5 × 3 =120
∴ Greatest 3-digit number which is divisible
by 8, 10 and 12 = 120 × 8 = 960

Olympiad Test : Playing With Numbers - 2 - Question 8

Which of the following number is not divisible by 9?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 8

3 + 8 + 7 + 4 + 5 + 9 = 36
9 + 0 + 4 + 8 + 0 + 6 = 27
7 + 5 + 8 + 9 +3 + 4 = 36
8 + 7 + 9 + 1 + 3 + 4 = 32
∴ 32 is not divisible by 9.
∴ 879134 is not divisible 9.

Olympiad Test : Playing With Numbers - 2 - Question 9

Find the smallest possible number which on adding 19 becomes exactly divisible by 28, 36 and 45.

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 9

LCM of 28, 36 and 45

∴ LCM of 28, 36 and 45
= 2 × 2 × 3 × 3 × 7 × 5
= 4 × 9 × 35
= 140 × 9
= 1260
∴ Required number = 1260 - 19 = 1241

Olympiad Test : Playing With Numbers - 2 - Question 10

Four bells toll at intervals 4, 7, 12 & 84 seconds. The bells toll together at 7 o’clock. How many times will they again toll together in 28 minutes?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 10

LCM of 4, 7, 12 and 84 =

= 2 × 2 × 3 × 7 = 84 seconds.
∴ The bells will toll together after 84 seconds.
∴ Number of times the bells will toll togetheer in 28 minutes

Olympiad Test : Playing With Numbers - 2 - Question 11

What is the least 5-digit number which is exactly divisible by 20, 25, 30?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 11

LCM of 20, 25, 30 is

∴ LCM of 20, 25, 30 = 2 × 5 × 2 × 5 × 3
= 300
∴ Ieast 5-digit number which is exactly divisible by 20, 25, 30 = 10200.

Olympiad Test : Playing With Numbers - 2 - Question 12

What is the maximum even multiple of 25 between 500 & 700?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 12

Multiples of 25 between 500 and 700 = 525, 550, 575, 600, 625, 650, 675.
∴ Required number = 650

Olympiad Test : Playing With Numbers - 2 - Question 13

Which of the following number is divisible by 8?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 13

694728 is divisible by 2 as well as 4.
∴ 694728 is divisible by 8.

Olympiad Test : Playing With Numbers - 2 - Question 14

Which of the following is divisible by 11?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 14

Difference = 13 – 13 = 0
∴ 65483 is divisible by 11.

Olympiad Test : Playing With Numbers - 2 - Question 15

What is sum of first five multiples of 23?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 15

Sum of first five multiples of 23 = 23 + 23 × 2 + 23 × 3 + 23 × 4 + 23 × 5
= 23 (1 + 2 + 3 + 4 + 5)
= 23 × 15
= 345

Olympiad Test : Playing With Numbers - 2 - Question 16

Which of the following statement is true?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 16

Numbers are divisible by 8 if the number formed by the last three individual digits is evenly divisible by 8.
344/8=43
hence, 1509344 is divisible by 8.
to verify:
1509344/8=188668.

Olympiad Test : Playing With Numbers - 2 - Question 17

In 467 * 381 replace * by which smallest digit to make it divisible by 3?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 17

467 * 381
The sum of digits of the above number
= 4 + 6 + 7 + * + 3 + 8 + 1
= 10 + 7 + * + 3 + 8 + 1
= 29 + *
If, ∗  = 1, then, the sum of digits will become 30.
∴ required number = 1

Olympiad Test : Playing With Numbers - 2 - Question 18

1870 is divisible by 22. Which two numbers nearest to 1870 are each divisible by 22?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 18

As 1870 is divisible by 2, Therefore subtracting and adding multiples of 22 will result in the numbers which are also divisible by 22.
1870 - 22 = 1848 & 1870 + 22 = 1892 are the numbers nearest to 1870, divisible by 22.

Olympiad Test : Playing With Numbers - 2 - Question 19

There are three heaps of rice weighing 120 kg, 144 kg and 204 kg. What is the maximum capacity of a bag so that the rice of each can be packed in exactly number of bags?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 19

Required capacity of bag
= HCF of 120, 144 and 204
= 12 kg

Olympiad Test : Playing With Numbers - 2 - Question 20

Four bells ring at intervals of 6, 8, 12 & 20 minutes. They ring simultaneously at 7 a.m. At what time will they ring together?

Detailed Solution for Olympiad Test : Playing With Numbers - 2 - Question 20

Required time = LCM of 6, 8, 12 and 20

∴ Required time = 2 × 2 × 3 × 2 × 5
= 120 minutes = 2 hr.
∴ bell will again ring together at (7 + 2)
= 9 a.m.
So at 9 A.M these 4 rings will ring together again.

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