MCQ: Unitary Method - 3 - SSC CGL MCQ

For 50 students, food is sufficient for 45 days
∴ For 1 student, food is sufficient for 45 x 50 days
and for 75 students, food is sufficient for (45 x 50)/75 days. i,e., for 30 days.

Let the required number of days be N.
Let men, More days (indirect proportion)
8 : 12 : : 24 : N
∴ N = (12 x 24)/8 = 36 days

Let the required number of days be N.
Remaining work = 1 - 5/8 = 3/8
Less work, Less days (Direct proportion)
5/8 : 3/8 : : 20 : N
⇒ N = (3/8) x 20 x 8/5
∴ N = 12 days

∵ Price of m articles = ₹ n
∴ Price of 1 article = ₹ n/m
∴ Price of 5 articles = ₹ 5n/m

∵ In 2/3h, 1 article is made.
∴ In 1h, 3/2 article are made.
∴ In 7¹/2 = 15/2h, 3/2 x 15/2 = 45/4 article are made.
∴ Required articles = 45/4 = 11²/4

Let the required length be L m.
More men, More length (Direct proportion)
Less days, Less length (Direct proportion)
Men 20 : 70
Days 6 : 3
∴ (20 x 6 x L) = (70 x 3 x 56)
∴ L = (70 x 3 x 56) / (20 x 6) = 98 m

Here M₁ = 30, D1 = 16,
T₁ = 9, M₂ = 36 and T₂ = 8
By formula, M₁D₁T₁ = M₂D₂T₂
30 x 16 x 9 = 36 x D2 x 8
∴ D₂ = (30 x 16 x 9)/(36 x 8) = 15 days

M₁ = 12 H₁ = 16, W₁ = 33600
M₂ = 18, H₂ = 12, W₂ = ?
By formula, M₁H₁/W₁ = M₂H₂/W₂
(12 x 16)/33600 = (18 x 12)/W₂ 
⇒ W₂ = (33600 x 18)/16
⇒ W₂ = 2100 x 8
∴ W₂ = 37800

∵ 22 men do the work in 16 days.
∴ 1 man will do the work in 16 x 22 days.
∴ 32 men will do the job in = (16 x 22)/32 days
∴ Required number of days = (16 x 22)/32 = 11 days

∵ For 1000 men, provision lasts for 48 days.
∴ For 1 men, provision lasts for (48 x 1000) days.
∴ For (1000 + 600) men, provision will last for (48 x 1000)/1600 days.
∴ Required number of days = (48 x 1000)/1600 = 3 x 10 = 30

Let the required number of flies be N.
More spiders, More flies (Direct proportion)
More time, More flies (Direct proportion)
∴ 10 x 10 x N = 200 x 200 x 10
∴ N = (200 x 200 x 10)/(10 x 10) = 4000 flies

∵ For 50 students, food is sufficient for 45 days.
∴ For 1 students, food is sufficient for 45 x 50 days.
∴ For 75 students, food is sufficient for (45 x 50)/75 days i.e., for 30 days .

3 men ≡ 6 boys
⇒ 1 man ≡ 2 boys
∴ 4 men + 4 boys ≡ 4 men + 2 men = 6 men
∵ 3 men can do a work in 18 days.
∴ 1 man can do a work in 18 x 3 days.
∴ Required number of days = (18 x 3)/6 = 9 days

Let the number of soldiers transferred be N
Now, the food would last for 25 days for (2000 - N) soldiers.
Less men, More days (indirect proportion)
25 : 20 : : 2000 : (2000 - N)
⇒ 2000 - N = (2000 x 20)/25
⇒ 2000 - N = 1600
∴ N = 2000 - 1600 = 400

Let the required number of days be N.
More men, Less days (indirect proportions)
Soldiers 90 : 72
Ratio 9 : 10
N = (72 x 54 x 10)/(90 x 9) = 48 days