MCQ: Coded Inequality- 2 - SSC CGL MCQ

Given statement: B > A; G < E > T; T ≥ B

On combining the statements,

G < E > T ≥ B > A

Checking Option A:

E > A is true.

Checking Option B:

T > A is true.

Checking Option C:

G > B is not true because of opposite sign between them.

Hence, Option C is correct.

Given statements: A ≥ B; J > E; O < F = A; B > J
On combining the statements:
O < F = A ≥ B> J > E
Checking Option A:
F > J is true.
Checking Option B:
A > E is true.
Checking Option D:
B > O is not true because of opposite sign between them.
Hence, Option D is correct.

Check options one by one
Checking Option A:
K = H < G ≤ I < J
Here, K < J is true but I ≥ H is not true because it make definite relation I < H.
Checking Option B:
K = H ≤ G ≤ I < J
Here, K < J and I ≥ H both follow.
Checking Option C:
K = H = G ≤ I < J
Here, K < J and I ≥ H both follow.
Hence, Option D is correct.

Check options one by one
Checking Option A: 75 = 66 ≤ 85 < 68 < 99 < 96
Here, 85 < 96 is true but 68 ≥ 75 is not true.
Checking Option B: 75 = 66 ≤ 85 ≤ 68 < 99 < 96
Here, 85 < 96 and 68 ≥ 75 both follow.
Checking Option C: 75 = 66 ≤ 85 > 68 < 99 < 96
Here, 85 < 96 and 68 ≥ 75 both doesn't follow.
Checking Option D: 75 = 66 ≤ 85 ≥ 68 < 99 < 96
Here, 85 < 96 and 68 ≥ 75 both doesn't follow.
Hence, Option B is correct.

Let us check each of the options one by one.
__ ≤   __ < __ >   __<

Option A: Here, W ≤ L < M > E < Q
So, Q > W is not true.
Thus option A does not follow.
Option B: Here, E ≤ W < L > Q < N
So, Q > W is not true.
Thus option B does not follow.
Option C: Here, H ≤ D < Q > M < W
So, Q > W is not true.
Thus, option C does not follow.
Hence, Option D is correct.

B ≥ U > O; K < B ___ Q ≤ C
On combining,
O < U ≤ B ____ Q ≤ C
Replacing the signs of option A in the given inequality,
O < U ≤ B > Q ≤ C
It doesn’t satisfy the given condition.
Replacing the signs of option B in the given inequality,
O < U ≤ B ≤ Q ≤ C
It satisfies the given condition.
Replacing the signs of option C in the given inequality,
O < U ≤ B ≥ Q ≤ C
It doesn’t satisfy the given condition.
Replacing the signs of option D in the given inequality,
O < U ≤ B < Q ≤ C
It doesn’t satisfy the given condition.
Hence, Option B is correct.

T = H ≥ S; N ? L ≤ S
On combining,
N ? L ≤ S ≤ H = T
Replacing the signs of option A in the given inequality,
N < L ≤ S ≤ H = T
It satisfies the given condition.
Replacing the signs of option B in the given inequality,
N ≤ L ≤ S ≤ H = T
It doesn’t satisfy the given condition.
Replacing the signs of option C in the given inequality,
N ≥ L ≤ S ≤ H = T
It doesn’t satisfy the given condition.
Replacing the signs of option A in the given inequality,
N > L ≤ S ≤ H = T
It doesn’t satisfy the given condition.
Hence, Option A is correct.

Combining statements: L > W ≥ Q = G; B ≤ G < K = M
Checking C1: B > L
Here, B ≤ G = Q ≤ W < L.
So, B < L is definitely true.
Therefore, B > L is not true.
Checking C2: M < W
Here, W ≥ Q = G < K = M.
We can clearly see that the signs between M and W get reversed therefore the relation between M and W cannot be formed.
Therefore, M < W is not true.
Hence, Option D is correct.

Combining statements: 124 > 343 = 149 ≥ 248; 149 < 328 ≤ 510; 510 > 320 < 620
Checking C1: 343 < 510
Here, 343 = 149 < 328 ≤ 510
The common sign between 343 and 510 is =, < and ≤.
Therefore, 343 < 510 is true.
Checking C2: 620 < 328
Here, 328 ≤ 510 > 320 < 620.
We can clearly see that the signs between 620 and 328 get reversed therefore the relation between 620 and 328 cannot be formed.
Therefore, 620 < 328 is not true.
Hence, Option A is correct.

Combining statements: M = N > P ≥ T; L > S ≥ M = A < B
Checking C1: T ≤ M
Here, M = N > P ≥ T.
So, T < M is definite true.
Therefore, T ≤ M is not true.
Checking C2: L < M
Here, L > S ≥ M.
So, L > M is definite true.
Therefore, L < M is not true.
Checking C3: S < B
Here, S ≥ M = A < B
We can clearly see that the signs between S and B get reversed therefore the relation between S and B cannot be formed.
Therefore, S < B is not true.
Hence, Option D is correct.

Combining statements: 1 ≤ 2; 3 = 4; 1 ≥ 4; 5 > 2
Checking C1: 3 = 1
Here, 3 = 4 ≤ 1.
The common sign between 3 and 1 are = and ≤.
So, 3 = 1 is not definately true.
Checking C2: 1 > 3
Here, 3 = 4 ≤ 1.
The common sign between 3 and 1 are = and ≤.
So, 1 > 3 is not definitely true.
Checking C3: 5 ≤ 2
Here, 5 > 2 is definite true.
So, 5 ≤ 2 is not true.
Conclusions I and II are complementary pairs.
Hence, Option D is correct.

By combining all the statements we get:
N ? P ≤ I ≤ T = H
Check options one by one
Checking Option A:
N < P ≤ I ≤ T = H, it satisfies the given condition.
Checking Option B:
N ≤ P ≤ I ≤ T = H, it doesn’t satisfy the given condition.
Checking Option C:
N > P ≤ I ≤ T = H, it doesn’t satisfy the given condition.
Checking Option D:
N ≥ P ≤ I ≤ T = H, it doesn’t satisfy the given condition.
Hence, Option A is correct.

Given statement: S ≥ D < P = F > O > Q = X > W
Checking C1: S > P
Here, S ≥ D < P, so no relation is possible because of opposite sign between them.
Therefore, C1 doesn’t follow.
Checking C2: F < X
Here, F > O > Q = X, so F > X is true.
So, F < X is not true.
Therefore, C2 doesn’t follow.
Checking C3: W < P
Here, P = F > O > Q = X > W, so W < P is true.
Therefore, C3 follows.
Hence, Option B is correct.

By combining all the statements we get:
S < L ≤ J > M ≥ K; R < M ≤ G
Checking C1: S < M
Here, S < L ≤ J > M, so no relation is possible because of opposite sign between them.
Therefore, C1 doesn’t follow.
Checking C2: G ≥ K
Here,  G ≥ M ≥ K, so G ≥ K is true.
Therefore, C2 follows.
Checking C3: M < L
Here, L ≤ J > M, so no relation is possible because of opposite sign between them.
Therefore, C3 doesn’t follow.
Hence, Option D is correct.

By combining all the statements we get:
M < L < N ≤ K < J; J > L > G
Checking C1: K > L
Here, M < L < N ≤ K < J, so K > L.
Therefore, C1 follows.
Checking C2: G > M
Here, M < L > G, so no relation is possible because of opposite sign between them.
Therefore, C2 doesn’t follow.
Hence, Option A is correct.