Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Number System Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- Find the maximum number of children among whom 150 pieces of chocolates and 225 pieces of fruits can be equally distributed?
- 85
- 75
- 65
- 55

Answer: (b) 75

Solution: To find the maximum number of children to distribute chocolates and fruits equally, we can find out the HCF of the number of chocolates and fruits. Hence HCF of 150 and 225.

Factorize the numbers into their prime numbers. $$ 150 = 2 \times 3 \times 5^2 $$ $$ 225 = 3^2 \times 5^2 $$ collect all the common factors with their minimum available power. $$ = 3 \times 5^2 $$ $$ HCF = 75 $$ Hence the maximum of 75 children will get equal pieces of chocolates and fruits.

Solution: To find the maximum number of children to distribute chocolates and fruits equally, we can find out the HCF of the number of chocolates and fruits. Hence HCF of 150 and 225.

Factorize the numbers into their prime numbers. $$ 150 = 2 \times 3 \times 5^2 $$ $$ 225 = 3^2 \times 5^2 $$ collect all the common factors with their minimum available power. $$ = 3 \times 5^2 $$ $$ HCF = 75 $$ Hence the maximum of 75 children will get equal pieces of chocolates and fruits.

- Four bells start to ring at intervals 5, 8, 10, and 12 seconds respectively. Find after what interval of time will they ring together?
- 100 seconds
- 110 seconds
- 120 seconds
- 125 seconds

Answer: (c) 120 seconds

Solution: To find the interval of time of four bells to ring together, we can find the LCM of 5, 8, 10, and 12. Here the LCM of 5, 8, 10, and 12 is 120. Hence after 120 seconds, the bells will ring together.

Solution: To find the interval of time of four bells to ring together, we can find the LCM of 5, 8, 10, and 12. Here the LCM of 5, 8, 10, and 12 is 120. Hence after 120 seconds, the bells will ring together.

- What is the largest size of tile which could be used for a rectangular terrace of 2.24 meters long and 3.12 meters wide?
- 5 cm.
- 6 cm.
- 7 cm.
- 8 cm.

Answer: (d) 8 cm.

Solution: First, we need to convert length into centimeters. So multiply 100 with the meter to convert it into cm. Hence the terrace is 224 cm long and 312 cm wide.

The HCF of 224 and 312 will be the maximum size of the tile.

HCF of 224 and 312.

Factorize the numbers into their prime numbers. $$ 224 = 2^5 \times 7 $$ $$ 312 = 2^3 \times 3 \times 13 $$ collect all the common factors with their minimum available power. $$ = 2^3 = 8 $$ $$ HCF = 8 $$ Hence 8 cm is the largest size of the tile.

Solution: First, we need to convert length into centimeters. So multiply 100 with the meter to convert it into cm. Hence the terrace is 224 cm long and 312 cm wide.

The HCF of 224 and 312 will be the maximum size of the tile.

HCF of 224 and 312.

Factorize the numbers into their prime numbers. $$ 224 = 2^5 \times 7 $$ $$ 312 = 2^3 \times 3 \times 13 $$ collect all the common factors with their minimum available power. $$ = 2^3 = 8 $$ $$ HCF = 8 $$ Hence 8 cm is the largest size of the tile.

- Find the least number of tiles required for a square room of 14 meters long and 10.20 meters wide?
- 3570
- 2270
- 3230
- 4128

Answer: (a) 3570

Solution: If the size of the tiles will be maximum then the required number of tiles will be least. First, we need to convert length into centimeters. So multiply 100 with the meter to convert it into cm. Hence the terrace is 1400 cm long and 1020 cm wide.

The HCF of 1400 and 1020 will be the maximum size of the tile.

HCF of 1400 and 1020.

Factorize the numbers into their prime numbers. $$ 1400 = 2^3 \times 5^2 \times 7 $$ $$ 1020 = 2^2 \times 3 \times 5 \times 17 $$ collect all the common factors with their minimum available power. $$ = 2^2 \times 5 $$ $$ HCF = 20 $$ Hence the maximum size of the tile will be 20 cm.

Now the least number of tiles required for a square room. $$ = \frac{1400 \times 1020}{20^2} $$ $$ = 3570 $$ Hence at least 3570 tiles required.

Solution: If the size of the tiles will be maximum then the required number of tiles will be least. First, we need to convert length into centimeters. So multiply 100 with the meter to convert it into cm. Hence the terrace is 1400 cm long and 1020 cm wide.

The HCF of 1400 and 1020 will be the maximum size of the tile.

HCF of 1400 and 1020.

Factorize the numbers into their prime numbers. $$ 1400 = 2^3 \times 5^2 \times 7 $$ $$ 1020 = 2^2 \times 3 \times 5 \times 17 $$ collect all the common factors with their minimum available power. $$ = 2^2 \times 5 $$ $$ HCF = 20 $$ Hence the maximum size of the tile will be 20 cm.

Now the least number of tiles required for a square room. $$ = \frac{1400 \times 1020}{20^2} $$ $$ = 3570 $$ Hence at least 3570 tiles required.

- Find the number between 300 and 400, which is exactly divisible by 6, 7, and 8?
- 312
- 322
- 336
- 345

Answer: (c) 336

Solution: The LCM of 6, 7, and 8.

Factorize the numbers into their prime factors. $$ 6 = 2 \times 3 $$ $$ 7 = 1 \times 7 $$ $$ 8 = 2^3 $$ collect all the distinct factors with their highest power. $$ LCM = 2^3 \times 3 \times 7 $$ $$ LCM = 168 $$ The multiple of 168 will also be exactly divisible by 6, 7, and 8. Hence the number 336 lies in between 300 and 400, which is exactly divisible by 6, 7, and 8.

Solution: The LCM of 6, 7, and 8.

Factorize the numbers into their prime factors. $$ 6 = 2 \times 3 $$ $$ 7 = 1 \times 7 $$ $$ 8 = 2^3 $$ collect all the distinct factors with their highest power. $$ LCM = 2^3 \times 3 \times 7 $$ $$ LCM = 168 $$ The multiple of 168 will also be exactly divisible by 6, 7, and 8. Hence the number 336 lies in between 300 and 400, which is exactly divisible by 6, 7, and 8.

Lec 1: Introduction to Number System
Lec 2: Factors of Composite Number
Questions and Answers-1
Lec 3: Basic Remainder Theorem
Questions and Answers-2
Lec 4: Polynomial Remainder Theorem
Questions and Answers-3
Questions and Answers-4
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Lec 5: LCM of Numbers
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Lec 6: HCF of Numbers
Questions and Answers-7
Questions and Answers-8
Lec 7: Divisibility Rules of Numbers
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