Practice With math operations - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   The sum of a pair of positive integer is 336 and their H.C.F. is 21. The number of such possible pairs is

(a)

(b)

(c)

(d)

Explanation:

Let the numbers be 21x and 21y

where x and y are prime to each other.

21x + 21y = 336

21 (x + y) = 336

x + y = $336/21$ = 16

∴ Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)


Q-2)   The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

1st number × 2nd number = L.C.M. × H.C.F

Let the larger number be x.

Smaller number = x – 2

First number × Second number = HCF × LCM

→ x (x – 2) = 24

→ $x^2$ – 2x – 24 = 0

→ $x^2$ – 6x + 4x – 24 = 0

→ x (x – 6) + 4 (x – 6) = 0

→ (x – 6) (x + 4) = 0

→ x = 6 because x ≠ – 4


Q-3)   The sum of two numbers is 84 and their HCF is 12. Total number of such pairs of number is

(a)

(b)

(c)

(d)

Explanation:

HCF = 12

∴ Numbers = 12x and 12y where x and y are prime to each other.

∴ 12x + 12y = 84

⇒ 12 (x + y) = 84

⇒ x + y = $84/12$ = 7

∴ Possible pairs of numbers satisfying this condition

= (1,6), (2,5) and (3,4). Hence three pairs are of required numbers.


Q-4)   If the product of three consecutive numbers is 210 then sum of the smaller number is :

(a)

(b)

(c)

(d)

Explanation:

2210
3105
535
7

210 = 2 × 3 × 5 × 7 = 5 × 6 × 7

∴ Required answer = 5 + 6 = 11


Q-5)   The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

(a)

(b)

(c)

(d)

Explanation:

HCF of two numbers = 27

Let the numbers be 27x and 27y

where x and y are prime to each other.

According to the question,

27x + 27y = 216

27 (x + y) = 216

x + y = $216/27$ = 8

Possible pairs of x and y = (1, 7) and (3, 5)

Numbers =(27, 189) and (81, 135)


Q-6)   The sum of two numbers is 36 and their H.C.F. is 4. How many pairs of such numbers are possible ?

(a)

(b)

(c)

(d)

Explanation:

HCF of two numbers = 4.

Hence, the numbers can be given by 4x and 4y

where x and y are co-prime.

Then, 4x + 4y = 36

4 (x + y) = 36

x + y = 9

Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)


Q-7)   Let x be the smallest number, which when added to 2000 makes the resulting number divisible by 12, 16, 18 and 21. The sum of the digits of x is

(a)

(b)

(c)

(d)

Explanation:

212, 16, 18, 21
26, 8, 9, 21
33, 4, 9, 21
1, 4, 3, 7

LCM = 2 × 2 × 3 × 4 × 3 × 7 = 1008

Multiple of 1008 = 2016

∴ Required number = 2016 – 2000 = 16 = x

∴ Sum of digits of x = 1 + 6 = 7


Q-8)   The sum of the H.C.F. and L.C.M of two numbers is 680 and the L.C.M. is 84 times the H.C.F. If one of the number is 56, the other is :

(a)

(b)

(c)

(d)

Explanation:

Let HCF be h and LCM be l.

Then, l = 84h and l + h = 680

→ 84h + h = 680

→ h = $680/85$ = 8

∴ l = 680 – 8 = 672

∴ Other number = ${672× 8}/56$ = 96


Q-9)   HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is :

(a)

(b)

(c)

(d)

Explanation:

Let the numbers be 7x and 7y

where x and y are co-prime.

Now, LCM of 7x and 7y = 7xy

∴ 7xy = 140 ⇒ xy = $140/7$ = 20

Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.

∴ Numbers are 28 and 35 which lie between 20 and 45.

∴ Required sum = 28 + 35 = 63.


Q-10)   The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Suppose 1st number is x then, 2nd number = 100 – x

∴ LCM × HCF = 1st number × 2nd number

→ 495 × 5 = x × (100 – x)

→ 495 × 5 = 100x – $x^2$

→ $x^2$ – 55x – 45x – 2475 = 0

→ (x – 45) (x – 55) = 0

→ x = 45 or x = 55

Then, difference = 55 – 45 = 10