IBPS PO Prelims 2018 Quantitative Aptitude Test-Day 1

Here for equation I
p² +23p + 130 = 0
⇒ (p + 13)(p + 10) = 0
⇒ p = -13, -10
for equation II
q² – 25 = 0
⇒ q² = 25
⇒ q = ± 5
So, ±5 > -13, -10

1. a²– 60a + 899 = 0
   ⇒ a² – 31a – 29a + 899 = 0
   ⇒ a(a – 31) – 29(a – 31) = 0
   ⇒ (a – 29)(a – 31) = 0
   Then, a = +29 or a = + 31
   b²– 71b +1240 = 0
   ⇒ b² – 40b – 31b +1240 = 0

⇒ b(b – 40) – 31(b – 40) = 0
⇒ (b – 31)(b – 40) = 0
Then, b = +31 or b = + 40
So, when b = +31, a = b for a = 31 and b > a for a = + 29
And when b = +40, b > a for a = +29 and b > a for a = +31
∴ So, we can observe that b ≥ a.

I. 7x – 3y = 13
Multiplying on both sides by 4, we get,
⇒ 28x – 12y = 52 —-(1)
II. 5x + 4y = 40
Multiplying on both sides by 3, we get,
⇒ 15x + 12y = 120 —-(2)
Adding equations 1 and 2, we get,
⇒ 43x = 172
⇒ x = +4 —-(3)
Substituting equation 3 in equation 2, we get,
⇒ 60 + 12y = 120
⇒ 12y = 60
⇒ y = +5
∴ We can observe that x < y.

1. x²– 80 = 244
   or, x² = 324
   ⇒ x = ±18
   y + 122 = 136
   or, y = 14
   Now, we can observe that →
   + 18 > 14, ∴ x > y-18 < 14, ∴ x < y
   So, no relation cannot be established between x and y.

1. 28a²– a – 2 = 0
   ⇒ 28a² – 8a + 7a – 2 = 0
   ⇒ 4a(7a – 2) + 1(7a – 2) = 0
   ⇒ (4a + 1) (7a – 2) = 0
   Then, a = (-1/4) or a = 2/7
   26b²– 27b + 7 = 0
   ⇒ 26b² – 14b – 13b + 7 = 0
   ⇒ 2b(13b – 7) – 1(13b – 7) = 0
   ⇒ (2b – 1) (13b – 7) = 0
   Then, b = (1/2) or b = (7/13)
   So, when a = (-1/4), a < b for b = (1/2) and a < b for b = (7/13)
   And when a = (2/7), a < b for b = (1/2) and a < b for b = (7/13)
   ∴ we can observe that a < b.

÷3-1, ÷3-1, ÷3-1, ÷3-1…

+0.5, +1.5, +0.5, +1.5, +0.5, +1.5

⇒ Original series: 1, 1, 2, 3, 4, 7
⇒ Sum of adjacent terms: 2, 3, 5, 7, 11 (Consecutive prime numbers)
∴ The next term = 13 – 7 = 6

The pattern of the series is as follows:
⇒ 3
⇒ 3 + 13 = 4
⇒ 4 + 23 = 12
⇒ 12 + 33 = 39
⇒ 39 + 43 = 103
⇒ 103 + 53 = ?
∴ ? = 228

*7-7, *6-6, *5-5, *4-4, *3-3, *2-2