**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 + 1

^{2}= 13
6 x 5 + 2

^{2}= 34
8 x 7 + 3

^{2}= 65
10 x 9 + 4

^{2}= 106
12 x 11 + 5

^{2}= 157
14 x 13 + 6

^{2}= 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 - 4

^{2}+ 1 = 2070
2070 - 5

^{2}+ 1 = 2046
2046 - 6

^{2}+ 1 = 2011
2011 - 7

^{2}+ 1 = 1963
1963 - 8

^{2}+ 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(Y

^{2}- 100)
6Y

^{2}- 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 1

^{st}vessel = n X 7/10 = 7n/10 liter
Amount of milk in 2

^{nd}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 = (

^{5}C_{2}x^{5}C_{3}) = 100
Committee of 5 member having 3 women
and 2 men = (

^{5}C_{3}x^{5}C_{2}) = 100
Committee of 5 member having 4 women
and 1 man = (

^{5}C_{4}x^{5}C_{1}) = 25
Committee of 5 member having 5 women
=

^{5}C_{5}= 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

63Y

^{2}= 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