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Hello Students,

We are sharing Data Interpretation Questions which are very important for  IBPS Clerk, IBPS PO, SBI Clerk, SBI PO, LIC AAO, Assitant, Bank Exams, SSC and Railway. Attempt these questions. 5 questions are asked in every bank exams. Previous year Question, Data Interpretation Questions Direction (1-10): In the following questions two equations numbered I and II are given. You have to solve both equations and give answer

1. I. x2 – 7x + 10 = 0

II. y2 + lly + 10 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

2. I x2 + 28x + 192 = 0

II. y2 + 16y + 48 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

3. I. 2x – 3y = – 3.5

II. 3x – 2y = – 6.5

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

4. I. x2 + 8x + 15 = 0

II. y2 + 11y + 30 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

5. I. x = √3136

II. y2 = 3136

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

6. I. x2 + 5x + 6 = 0

II. y2 + 7y + 12 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

7. I. x2 + 20 = 9x

II. y2 + 42 = 13y

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

8. I. 2x + 3y = 14

II. 4x + 2y = 16

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

9. I. x = √625

II. x = √676

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

10. I. x2 + 4x + 4 = 0

II. y2 – 8y + 16 = 0

(a) x > y

(b) x ≥ y

(c) x < y

(d) x ≤ y

(e) x = y or the relationship cannot be established

I. x2 – 7x + 10 = 0

x2 – 5x – 2x + 10 = 0

x(x – 5) – 2(x – 5) = 0

(x – 2) (x – 5) = 0

x = 2 or 5

II. y2 + 11y + 10 = 0

y2 + 10y + y + 10 = 0

y(y + 10) + 1(y + 10) = 0

(y + 1) (y + 10) = 0

y = – 1 or – 10

Here, x > y

I. x2 + 28x + 192 = 0

x2 + 16x + 12x + 192 = 0

x(x + 16) + 12(x + 16) = 0

(x + 12) (x + 16) = 0

x = – 12 or –16

II. y2 + 16y + 48 = 0

y2 + 12y + 4y + 48 = 0

y (y + 12) + 4 (y + 12) = 0

(y + 12) (y + 4) = 0

y = – 12 or – 4

Here, x ≤ y

2x – 3y = – 3.5 … (i)

3x – 2y = – 6.5 … (ii)

By equation (i) × 2 – equation (ii) × 3, we have

-5x = 12.5

x = -12.5/5 = –2.5

From equation (i)

2 × (–2.5) – 3y = – 3.5

3y = – 5 + 3.5

3y = – 1.5

y = -0.5

Here, x < y

I. x2 + 8x + 15 = 0

x2 + 5x + 3x + 15 = 0

x(x + 5) + 3(x + 5) = 0

(x + 5) (x + 3) = 0

x = –5 or –3

II. y2 + 11y + 30 = 0

y2 + 6y + 5y + 30 = 0

y (y + 6) + 5 (y + 6) = 0

(y + 5) (y + 6) = 0

y = – 5 or – 6

Here, x ≥ y

I. x = √3136

x = 56

II. y2 = 3136

y = √3136

x = ± 56

Here, x ≥ y

I. x2 + 5x + 6 = 0

x2 + 2x + 3x + 6 = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x = -2, -3

II. y2 + 7y + 12 = 0

y2 + 4y + 3y + 12 = 0

y(y + 4) + 3(y + 4) = 0

(y + 4)(x + 3) = 0

x = -3, -4

Here, x ≥ y

I. x2 – 9x + 20 = 0

x2 – 5x – 4x + 20 = 0

x(x – 5) – 4(x – 5) = 0

(x – 4)(x – 5) = 0

x = 4, 5

II. y2 – 13y + 42 = 0

y2 – 6y – 7y + 42 = 0

y(y – 6) -7(y – 6) = 0

(y – 6)(y – 7) = 0

y = 6, 7

Here, x < y

2 x + 3 y = 14 …(i)

4 x + 2 y =16 …(ii)

By equation (I) × 2 – equation II.

4 x + 6y – 4 x – 2 y = 28 – 16

4y =12

y = 3

From equation I,

2 x + 3× 3 = 14

2 x = 14 – 9 = 5

x = 5/2

Here, x < y

I. x = √625 = 25

II. y = √676 = 26

Here, x < y

I. x2 + 4x + 4 = 0

(x + 2)2 = 0

x = – 2

II. y2 – 8y + 16 = 0

(y – 4)2 = 0

y = 4

Here, x < y