Question

If 52272 = p²×q³×r⁴ where p, q and r are prime numbers, then the value of (2p+ q-r) is
(1) 23
(2) 29
(3) 21
(4) 22

Answer 1

Answer:

23

Step-by-step explanation:

If 52272 = p²×q³×r⁴ where p, q and r are prime numbers, then the value of (2p+ q-r) is  

(1) 23

(2) 29

(3) 21

(4) 22

52272 = 2 * 2 * 2 * 2  *  3 * 3 * 3  *  11 * 11

=> 52272 = 2⁴ * 3³  * 11²

=> 52272 = 11² *  3³ * 2⁴

Comparing with

52272 = p²×q³×r⁴

p = 11

q = 3

r = 2

2p + q - r  = 2 * 11 + 3 - 2 = 23

Answer 2

The value of (2p + q - r) is (1) "23".

Step-by-step explanation:

We have,

$52272 =p^{2} \times q^{3} \times r^{4}$    ... (1)

∴ 52272 = 2 × 2 × 2 × 2 × 3 × 3 × 3  ×  11 × 11

$= 2^{4} \times 3^{3} \times 11^{2}$

$= 11^{2} \times 3^{3} \times 2^{4}$                   ... (2)

Equating equations (1) and (2), we get

$p^{2} \times q^{3} \times r^{4}= 11^{2} \times 3^{3} \times 2^{4}$

∴ p = 11, q = 3 and  r = 2

∴ 2p + q - r

= 2 × 11 + 3 - 2

= 22 + 3 - 2

= 23

Hence, the value of (2p + q - r) is (1) 23.