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1.

1). A, C2). B, D3). B, A4). A, B

Answer»

SUPPOSE total work = 120 units (LCM of 20, 24 & 60)

⇒ Efficiency of (Shivam and Pankaj) = 120/20 = 6

⇒ Efficiency of (Mridul and Pankaj) = 120/24 = 5

⇒ Efficiency of Pankaj = 120/60 = 2

⇒ Efficiency of Mridul = 5 – 2 = 3

⇒ Efficiency of Shivam = 6 – 2 = 4

Given that:

Shivam started the work and WORKED for 12 days, then Mridul worked for 14 days, then Pankaj worked for x days,

⇒ 4 × 12 + 3 × 14 + 2x = 120

⇒ 2x = 120 – 90 = 30

⇒ x = 15

Quantity 1:

A can do the work in 20 days and B can do the same work in 20 days (15 + 5).

⇒ Number of days taken by A and B together to complete the work = 10 days

Quantity 2:

P can do some work in 10 (15 – 5) days while Q can do the same work in 15 days,

Suppose total work = 30 units (LCM of 10 & 15)

⇒ Efficiency of P = 30/10 = 3

⇒ Efficiency of Q = 30/15 = 2

⇒ Number of days taken by P and Q together to complete the work = 30/5 = 6 days

Quantity 3:

VALUE of x = 15

∴ Quantity 1 > Quantity 2 < Quantity 3
Discussion