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#### Information

**Name: **Maths Practice Test – 1

**Subject: **Mathematics

**Topic: **Basic ( Calculation Based)

**Questions: **12 Objective type

**Time Allowed: **20 Minutes

**Important for: **Class 7, 8 , 9 , 10, 11 & 12 students, SSC CHSL, SSC GD, Stenographer Railway, CRPF, BSF, Army, Police, CTET, State TET, Polytechnic, Engineering & Medical Entrance Exams.

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- Review

- Question 1 of 12
##### 1. Question

1 pointsIf 12 man can do a piece of work in 36 days. In how many days 18 men can do the same work?

Correct12 men can do a work in 36 days.

18 men can do the work in (12/18)×36 = 24 days.Note: If the number of man is increased, the number of days to finish the work will decrease.

Incorrect12 men can do a work in 36 days.

18 men can do the work in (12/18)×36 = 24 days.Note: If the number of man is increased, the number of days to finish the work will decrease.

- Question 2 of 12
##### 2. Question

1 pointsA man plants his orchid with 5625 trees and arranges them so that there are as many rows as there are trees in a row. How many rows are there?

CorrectLet ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say x

^{2}= 5625

∴x = 75

∴There are 75 trees in a row and 75 rows are arranged.IncorrectLet ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say x

^{2}= 5625

∴x = 75

∴There are 75 trees in a row and 75 rows are arranged. - Question 3 of 12
##### 3. Question

1 pointsWhat least number should be multiply with number 678, so that the product will become a perfect square?

CorrectFirst find the factors of 678.

678 = 2 × 3 × 113

Now, 678 has factors as shown above, ‘2’ ones, ‘3’ ones and ‘113’ ones only. So when one more factor of each factor like ‘2’, ‘3’ and ‘109’ is used, then it becomes a perfect square.

678 ×2 ×3 × 113 = (2 × 2) × (3 × 3) × (113 × 113)

The least number required is ‘679’.

IncorrectFirst find the factors of 678.

678 = 2 × 3 × 113

Now, 678 has factors as shown above, ‘2’ ones, ‘3’ ones and ‘113’ ones only. So when one more factor of each factor like ‘2’, ‘3’ and ‘109’ is used, then it becomes a perfect square.

678 ×2 ×3 × 113 = (2 × 2) × (3 × 3) × (113 × 113)

The least number required is ‘679’.

- Question 4 of 12
##### 4. Question

1 pointsIf a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.

CorrectLet the number be ‘x’, say

∴x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’

= 37 × 8k + 37 × 2 + 1

= 37(8k + 2) + 1

Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.

IncorrectLet the number be ‘x’, say

∴x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’

= 37 × 8k + 37 × 2 + 1

= 37(8k + 2) + 1

Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.

- Question 5 of 12
##### 5. Question

1 pointsIf m and n are two whole numbers and if m

^{n}= 25. Find n^{m}, given that n ≠ 1.Correctm

^{n}= 25 = 5^{2}

∴ m = 5; n = 2

∴ n^{m}= 2^{5}= 32Incorrectm

^{n}= 25 = 5^{2}

∴ m = 5; n = 2

∴ n^{m}= 2^{5}= 32 - Question 6 of 12
##### 6. Question

1 pointsFind the number of prime factors of 6

^{10}× 7^{17}× 55^{27}Correct6

^{10}× 7^{17}× 55^{27}= 2^{10}× 3^{10 }× 7^{17 }× 5^{27}× 11^{27}

∴The number of prime factors = the sum of all the indices viz., 10 + 10 + 17 + 27 + 27 = 91Incorrect6

^{10}× 7^{17}× 55^{27}= 2^{10}× 3^{10 }× 7^{17 }× 5^{27}× 11^{27}

∴The number of prime factors = the sum of all the indices viz., 10 + 10 + 17 + 27 + 27 = 91 - Question 7 of 12
##### 7. Question

1 pointsIf a number when divided by 54 gives a remainder 45, find the remainder when 18 divides the same number.

CorrectLet the number be ‘x’, say

∴x = 54k + 45, where ‘k’ is quotient when ‘x’ is divided by ‘54’

= 18 × 3k + 18 × 3 + 9

= 18(3k + 3) + 9

Hence, the remainder is ‘9’ when the number ‘x’ is divided by 18.

IncorrectLet the number be ‘x’, say

∴x = 54k + 45, where ‘k’ is quotient when ‘x’ is divided by ‘54’

= 18 × 3k + 18 × 3 + 9

= 18(3k + 3) + 9

Hence, the remainder is ‘9’ when the number ‘x’ is divided by 18.

- Question 8 of 12
##### 8. Question

1 pointsIf m and n are two whole numbers and if m

^{n}= 3125. Find n^{m}, given that n ≠ 1.Correctm

^{n}= 3125 = 5^{5}

∴ m = 5; n = 5

∴ n^{m}= 5^{5}= 3125Incorrectm

^{n}= 3125 = 5^{5}

∴ m = 5; n = 5

∴ n^{m}= 5^{5}= 3125 - Question 9 of 12
##### 9. Question

1 pointsFind the number of prime factors of 15

^{3}× 12^{4}× 17^{5}× 19^{2}.Correct15

^{3}× 12^{4}× 17^{5}× 19^{2}= (3 × 5)^{3}× (2^{2}×3)^{4}× 17^{5}× 19^{2}

= 2^{8}× 3^{7}× 5^{3}× 17^{5}× 19^{2}

∴The number of prime factors = the sum of all the indices viz., 8 + 7 + 3 + 5 + 2 = 25Incorrect15

^{3}× 12^{4}× 17^{5}× 19^{2}= (3 × 5)^{3}× (2^{2}×3)^{4}× 17^{5}× 19^{2}

= 2^{8}× 3^{7}× 5^{3}× 17^{5}× 19^{2}

∴The number of prime factors = the sum of all the indices viz., 8 + 7 + 3 + 5 + 2 = 25 - Question 10 of 12
##### 10. Question

1 pointsFour prime numbers are given in ascending order of their magnitudes, the product of the first three is 385 and that of the last three is 1001. Find the largest of the given prime numbers.

CorrectThe product of the first three prime numbers = 385

The product of the last three prime numbers = 1001

In the above products, the second and the third prime numbers occur in common.

∴ The product of the second and third prime numbers = HCF of the given products.

HCF of 385 and 1001 = 77

∴Largest of the given primes = 1001 / 77

= 13

IncorrectThe product of the first three prime numbers = 385

The product of the last three prime numbers = 1001

In the above products, the second and the third prime numbers occur in common.

∴ The product of the second and third prime numbers = HCF of the given products.

HCF of 385 and 1001 = 77

∴Largest of the given primes = 1001 / 77

= 13

- Question 11 of 12
##### 11. Question

1 pointsFour prime numbers are given in ascending order of their magnitudes, the product of the first three is 273 and that of the last three is 1729. Find the largest of the given prime numbers.

CorrectThe product of the first three prime numbers = 273

The product of the last three prime numbers = 1729

In the above products, the second and the third prime numbers occur in common.

∴ The product of the second and third prime numbers = HCF of the given products.

HCF of 273 and 1729 = 91

∴Largest of the given primes = 1729 / 91

= 19

IncorrectThe product of the first three prime numbers = 273

The product of the last three prime numbers = 1729

In the above products, the second and the third prime numbers occur in common.

∴ The product of the second and third prime numbers = HCF of the given products.

HCF of 273 and 1729 = 91

∴Largest of the given primes = 1729 / 91

= 19

- Question 12 of 12
##### 12. Question

1 pointsWhat least number should be multiply with number 882, so that the product will become a perfect square?

CorrectFirst find the factors of 882.

882 = 2 × 3 × 3 × 7 × 7

Now, 882 has factors as shown above, ‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only once. So when one more factor ‘2’ is used, then it becomes a perfect square.

882 × 2 = (2 × 2) × (3 × 3) × (7 × 7)

The least number required is ‘2’

IncorrectFirst find the factors of 882.

882 = 2 × 3 × 3 × 7 × 7

Now, 882 has factors as shown above, ‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only once. So when one more factor ‘2’ is used, then it becomes a perfect square.

882 × 2 = (2 × 2) × (3 × 3) × (7 × 7)

The least number required is ‘2’