Quantitative Aptitude Mains Level || 03 – 12 – 2018

Mentor for Bank Exams

Dear Aspirants,
Welcome to Mentor for Bank Exams. Here is the Quantitative Aptitude Quiz to help you practice with the best of latest pattern questions for the upcoming IBPS PO, IBPS Clerk and other bank and Insurance Exams.

Directions (1 – 5): Find the wrong term in the series given below.
1. 131, 190, 241, 308, 383, 466
a) 383
b) 241
c) 466
d) 190
e) 308
Answer: D)
Explanation:
The pattern is as follows:
10 x 13 + 1 = 131
12 x 15 + 2 = 182
14 x 17 + 3 = 241
16 x 19 + 4 = 308
18 x 21 + 5 = 383
20 x 23 + 6 = 466

2. 13, 26, 65, 106, 157, 218
a) 106
b) 218
c) 26
d) 65
e) 157
Answer: C)
Explanation:
The pattern is as follows:
4 x 3 + 12 = 13
6 x 5 + 22 = 34
8 x 7 + 32 = 65
10 x 9 + 42 = 106
12 x 11 + 52 = 157
14 x 13 + 62 = 218

3. 9, 27.5, 139.0, 971.5, 8783.0, 96617.5
a) 8783.0
b) 139.0
c) 96617.5
d) 971.5
e) 27.5
Answer: D)
Explanation:
The pattern is as follows:
9 x 3 + 0.5 = 27.5
27.5 x 5 + 1.5 = 139.0
139.0 x 7 + 2.5 = 975.5
975.5 x 9 + 3.5 = 8783.0
8783.0 x 11 + 4.5 = 96617.5

4. 2085, 2078, 2046, 2011, 1963, 1900
a) 2078
b) 1900
c) 2046
d) 1963
e) 2011
Answer: A)
Explanation:
The pattern is as follows:
2085 - 42 + 1 = 2070
2070 - 52 + 1 = 2046
2046 - 62 + 1 = 2011
2011 - 72 + 1 = 1963
1963 - 82 + 1 = 1900

5. 31, 40, 60, 93, 147, 206
a) 40
b) 147
c) 206
d) 93
e) 60
Answer: B)
Explanation:
The pattern is as follows:
31 + 9 = 40
40 + 9 + 11 = 60
60 + 9 + 11 + 13 = 93
93 + 9 + 11 + 13 + 15 = 141
141 + 9 + 11 + 13 + 15 + 17 = 206

Directions (6 – 10): In the following questions, two statements numbered I and II are given. On solving them, we get quantities I and II respectively. Solve for both the quantities and choose the correct option.
6. Quantity I: A boat goes from point P to Q downstream having a distance 120 m and comeback to point T which is in the middle of P and Q in 6 secs then what is the speed of boat(km/hr) if the speed of the stream is 10 m/sec?
Quantity II: The ages of Raju and Rohan 5 years ago is in the ratio 3:2 and after 15 years ratio of ages of Raju to Rohan is 17:13 then what is sum of the ages (years) of Raju and Rohan 9 years from now?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or the relation can't be determined
Answer: E)
Explanation:
Quantity I:
Let speed of boat be Y m/sec.
Then,
Upstream speed = (Y - 10) m/sec.
Downstream speed = (Y + 10)
Thus,
120/(Y + 10) + 60/(Y - 10) = 6
120Y - 1200 + 60Y + 600 = 6(Y + 10)(Y - 10)
180Y -600 = 6(Y2 - 100)
6Y2 - 180Y = 0
6Y(Y - 30)
Either Y = 0 or Y = 30
As speed of the boat cannot be zero during the journey.
Speed of boat = 30 m/sec.
Required speed = (30 x 18/5) = 108 km/hr
Quantity II:
Let ages of Raju and Rohan 5 years ago be 3Y and 2Y respectively.
Then,
(3Y + 20)/(2Y + 20) = 17/13
39Y + 260 = 34Y + 340
5Y = 80
Y = 16
Present age of Raju = (3 x 16) + 5 = 53 years
Present age of Rohan = (2 x 16) + 5 = 37
Required sum = (53 + 9) + (37 + 9) = 108 years
Hence, Quantity I = Quantity II

7. Quantity I: Milk and water is filled into two vessels of same capacity in which water and milk are in the ratio of 3 : 7 and 2 : 5. If the mixtures from both the vessels are poured into another container then what will be the ratio of milk and water in the new mixture?
Quantity II: There is a mixture of 85 liter in which milk and water are in the ratio of 7 : 1. If 15 liters of water is poured into the mixture then what will the ratio of milk and water in the resulting mixture?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or the relationship cannot be determined
Answer: B)
Explanation:
Quantity - I
Let the capacity of each vessel = n liter
Amount of milk in 1st vessel = n X 7/10 = 7n/10 liter
Amount of milk in 2nd vessel = n X 5/7 = 5n/7 liter
Total amount of milk in both the vessel = (7n/10 + 5n/7) liter = 99n/70 liter
Amount of water in the container after mixing = 2n - 99n/70 = 41n/70 liter
Therefore, ratio of milk and water in the resulting mixture = 99n/70 : 41n/70
=99 : 41
Quantity - II
Amount of water in the mixture = (85 X 1/8) liter = 10 5/8 liter
Amount of water in the mixture after pouring 15 liters in it = (15 + 10 5/8) liter
=25 5/8 liter
amount of milk in the resulting mixture = (100 - 25 5/8) liter = 74 3/8 liter
Hence required ratio = 74 3/8 25 5/8 = 595 205 = 119 41
Hence, quantity II is higher that quantity 1.

8. Quantity I: In how many ways a committee of 5 members can be formed from 5 men and 5 women having at least 2 women?
Quantity II: What is length (in meters) of train P which crosses a platform of length 280 meter and crosses train Q having length 360 meter running at a speed of 30 m/sec in opposite direction in 25.3 sec and 11.72 sec respectively?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or the relation can't be determined.
Answer: E)
Explanation:
Quantity I:
Committee of 5 member having 2 women and 3 men = (5C2 x 5C3) = 100
Committee of 5 member having 3 women and 2 men = (5C3 x 5C2) = 100
Committee of 5 member having 4 women and 1 man = (5C4 x 5C1) = 25
Committee of 5 member having 5 women = 5C5 = 1
Required possible number = (100 + 100 + 25 + 1) = 226
Quantity II:
We have:
Let the length of train P be Y meter
Length of platform crossed by train P = 280 meter
Speed of train P = (Y + 280)/25.3
Length of train Q = 360 meter
Speed of train Q = 30 m/sec
Relative speed of train P to Q = ((Y + 280)/25.3 + 30)
((Y + 280)/25.3 + 30) = (Y + 360)/11.72
(Y + 1039)/25.3 = (Y + 360)/11.72
11.72Y + 12177.08 = 25.3Y + 9108
13.58Y = 3069.08
Y = 226 meter
Hence, Quantity I = Quantity II

9. Quantity I: In a circle of radius 84 cm, an arc subtends an angle of 108° at the center, then what is the length (cm) of the arc?
Quantity II: What is length (m) of a rectangular field having cost of flooring at the rate of ₹4.8/m2 is ₹68040 and ratio of the length to breadth of the field is 9:7?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or the relation cannot be determined
Answer: A)
Explanation:
Quantity I:
Radius = 84 cm
Angle = 108°
Length of the Arc = 2ϖRθ/360° = (2 x 22/7 x 84 x 108/360)
Length of the Arc = 158.4 cm
Quantity II:
Let length of the field be 9Y and Breadth be 7Y
Area = (L x B) = 68040/4.8 = 14175
By putting values of length and breadth we get:
(9Y x 7Y) = 14175
63Y2 = 14175
Y = 15
Then, length = (15 x 9) = 135 m
Hence, Quantity I > Quantity II

10. Quantity I: Rajeev invested a certain amount at the rate of 12% compounded annually and obtained a compound interest of ₹5215.2 at the end of two years then what is total simple interest earned at the end of 5 years on same amount if rate is increased by 3%?
Quantity II: Raghu sold a mobile phone at a 12% profit. Had he sold it for ₹738 more he would get a profit of 18% then what was selling price for a profit of 25%?
a) Quantity I > Quantity II
b) Quantity I < Quantity II
c) Quantity I ≥ Quantity II
d) Quantity I ≤ Quantity II
e) Quantity I = Quantity II or the relation can't be determined.
Answer: E)
Explanation:
Quantity I:
Let initial sum be ₹P.
Rate = 12%
Time = 2 years
CI = P x ((1 + R/100)2 - 1)
P x ((1 + R/100)2 - 1) = 5215.2
P x ((1 + 12/100)2 - 1) = 5215.2
P x 2544/10000 = 5215.2
P = 5215.2 x 10000/2544
P = ₹20500
Again,
Rate = (12 + 3) = 15%
Time = 5 years
SI = (P x R x T)/100
SI = (20500 x 15 x 5)/100 = 15375
Quantity II:
Let cost price of mobile be ₹Y
Then,
1.18Y - 1.12Y = 738
0.06Y = 738
Y = 738/0.06 = 12300
Required selling price = 12300 x 125/100 = ₹15375
Hence, Quantity I = Quantity II