Online & Free Quant Quiz for IBPS PO & CLERK – Part – 61
Directions (1-5): Study the table given below and answer the following Question
Q1. Find the average no. of Females in HR department together?
(a) 54
(b) 46
(c) 49
(d) 50
(e) 52
Q2. Females in the HR dept of company C is what % more than male in HR department of company A?
(a) 250%
(b) 200%
(c) 100%
(d) 300%
(e) 150%
Q3. If total no. of employee in E is 25% more than D and no. of employee in HR dept is same as in company C, then employee other than HR dept in company E is what % of other dept employee in company B.
(a) 60%
(b) 80%
(c) 75%
(d) 50%
(e) 55%
Q4. Find the difference between males of HR dept in company C and D together and females of HR dept in company B and C together?
(a) 36
(b) 42
(c) 48
(d) 40
(e) 30
Q5. Find the average no. of employee other than HR dept. in A, B and C together?
(a) 280
(b) 270
(c) 220
(d) 300
(e) 240
S1. Ans. (c)
Sol. Average no. of females in HR dept =
80 × (75/100) + 50 × (80/100) + 100 × (60/100) + 60 × (60/100) / 4
= (60 + 40 + 60 + 36) / 4 = 196 / 4 = 49
S2. Ans. (b)
Sol. Females in company C (HR) = 100 × (60/100) = 60
Males in company A (HR) = 80 × (25/100) = 20
Difference = 60 – 20 = 40
∴ % = (40/20) × 100 = 200% more
S3. Ans. (c)
Sol. Total employee in E = 200 × (125/100) = 250
∴ Employee of HR dept in E = 100
∴ Other employee = 150
∴ % of other employee = 150 × (100/200) = 75%
S4. Ans. (a)
Sol. Males in HR dept in C and D = 100 × (40/100) + 60 × (40/100) = 40 + 24 = 64
Females in HR dept of B and C = 50 × (80/100) + 100 × (60/100) = 100
∴ Difference = 100 – 64 = 36
S5. Ans. (e)
Sol. Average of A, B, C = (220 + 200 + 300) / 3 = 720 / 3 = 240
Directions (6-10): Solve the given quadratic equations and mark the correct option based on your answer-
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or there is no relationship
Q6. (i) x² = 81 (ii) y² – 18y + 81 = 0
Q7. (i) 4x² – 24x + 32 = 0 (ii) y² – 8y + 15 = 0
Q8. (i) x² – 21x + 108 = 0 (ii) y² – 17y + 72 = 0
Q9. (i) x² – 11x + 30 = 0 (ii) y² – 15y + 56 = 0
Q10. (i) x³ = 512 (ii) y² = 64
S6. Ans. (d)
Sol. x² = 81
x = ± 9
Y² – 18y + 81 = 0
(y – 9)² = 0
∴ y = 9, 9
∴ x ≤ y
S7. Ans. (e)
Sol. 4x² – 24x + 30 = 0
4x² – 16x – 8x + 32 = 0
4x (x – 4) –8 (x–4) = 0
x = 4,
2 y² – 8y + 15 = 0
y² – 5y – 3y + 15 = 0
y(y – 5)–3 (y – 5) = 0
∴ y = 5, 3
∴ No relation exists
S8. Ans. (c)
Sol.x² – 21x + 108 = 0
x² – 9x – 12x + 108 = 0
x(x – 9) – 12 (x – 9) = 0
x = 9, 12
y² – 17y + 72 = 0
∴ y² – 8y – 9y + 72 = 0
y (y – 8) – 9 (y – 8) = 0
∴ y = 8, 9
∴ x ≥ y
S9. Ans. (b)
Sol. x² – 11x + 30 = 0
x² – 6x – 5x + 30 = 0
∴ x(x – 6) – 5(x – 6) = 0
x = 6, 5
y² – 15y + 56 = 0
y² – 7y – 8y + 56 = 0
y (y – 7) – 8 (y – 7) = 0
∴ y = 7, 8
∴ x < y
S10. Ans. (c)
Sol. x³ = 512
x = 3√512 = 8
y² = 64
y = √64 = ± 8
∴ x ≥ y
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