Directions (1-5): The following line graph shows the number of reams (packets of A4 size paper) in terms of percentage used by three departments of Career Power i.e. Print dept., HR dept. and DTP dept. Provided that total number of reams used per month is 1200 and it remains consistent for all the months. There is no other department using these reams.
Q1. Find the difference in total number of reams used by print dept. from January to March and that of by DTP dept. from April to May?
Q2. In May, Babu working in Print dept. used 25% of the reams from which he wastedreams. Find the number of reams not wasted by Babu.
Q3. In July HR dept. demanded 25% more reams than previous month by stationary supervisor Mr. Vinod, which he denied. But he provided them the reams
less than that provided to print dept. in June. Find the ratio of number of reams demanded and actually provided to HR dept.
Reams provided = 432 - 72 = 360
∴Required ratio = 375 ∶ 360 = 25:24
Q4. By approximately what percent the total number of reams used by DTP dept. is less than that of by print team throughout the given months?
Q5. Total number of reams used by print dept. DTP dept. and HR dept. in February, April and June respectively is what percent of total reams used by HR dept. in given 6 months ?
Directions (6-10): What should come in the place of the question mark (?) in following number series problems?
Q6.12, 12, 18, 45, 180, 1080, ?
∴ ? = 1080 ×8.5 = 9180
Q7. 444, 467, 513, 582, 674, 789, ?
∴? = 789 + 138 = 927
Q8.1, 16, 81, 256, 625, 1296, ?
Q9.23, 25, 53, 163, 657, 3291, ?
∴? = 3291 × 6 + 7 = 19753
Q10. 13, 13, 65, 585, 7605, 129285, ?
∴? = 129285 × 21 = 2714985
Directions (11-15): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
Q11. I. 49x² + 84x – 64 = 0
II. 25y² - 100y + 64 = 0
Q12. I. 4x² - 33x + 63 = 0
II. 10y² - 113y + 318 = 0
Q13. I. 3x + 4y = 24
II. 2x + 3y = 17
II. 2x + 3y = 17
Solving (I) & (II)
x = 4, y = 3
x > y
Q15. I. x² – 15x + 56 = 0
II. y² – 13y + 42 = 0
x² – 7x – 8x + 56 = 0
x (x – 7) – 8 (x – 7)
x = 7, 8
II. y² – 13y + 42 = 0
y² – 6y – 7y + 42 = 0
y (y – 6) – 7 (y – 6) = 0
y = 7, 6
x ≥ y