Aptitude Questions (Arithmetic) for IBPS Clerk Mains

Aptitude Questions (Arithmetic) for IBPS Clerk Mains
Aptitude Questions (Arithmetic) for IBPS Clerk Mains:
Dear Readers, Important Practice Aptitude Questions for Upcoming IBPS Clerk V Mains Exam was given here with Explanations. Aspirants those who are preparing for the examination can use this.

Directions (Q.1-10): Study the following information carefully and the answer given questions:
 
1). The price of sugar falls by 12%. How many quintals can be bought for the same money which was sufficient to buy 44 quintals at the higher price?
a)    54
b)    50
c)    38
d)    48
e)    None of these
2). The distance between two stations A and B is 220 km. Two motorcyclists start simultaneously from A and B in opposite directions and the distance between them after 2 hours is 30 km. The faster motorcyclist’s speed is 5 km/hr more than the slower motorcyclist’s. Find the ratio of the faster motorcyclist’s speed to the slower motorcyclist’s.
a)    9 : 5
b)    10 : 11
c)    10 : 9
d)    5 : 8
e)    None of these
3). A field can be plouged in 16 days. If 18 more hectares of land is plouged daily, the work will be finished in 10 days. Find the area of the field.
a)    480 hectares
b)    520 hectares
c)    440 hectares
d)    500 hectares
e)    None of these
4). An article Rs. 425 is marked to be sold at a price which gives a profit of 20%. What will be its selling price in a sale when 15% is taken off the marked price?
a)    Rs. 442
b)    Rs. 430
c)    Rs. 438
d)    Data inadequate
e)    None of these
5). How much pure alcohol can be added to 700 ml of an 18% solution to make its strength 30%?
a)    60 ml
b)    100 ml
c)    140 ml
d)    120 ml
e)    None of these
6). A ladder 12 m long reaches a point 8 m below the top of a building. At the foot of the ladder the elevation of the top of the building is 60°. Find the approx height of the building.
a)    18 m
b)    21 m
c)    16 m
d)    Data inadequate
e)    14 m
7). Find the value of x in the equation
(x³ + 3x / 3x² + 1) = 1241/217
a)    17
b)    19
c)    13
d)    27
e)    None of these
8). A hemispherical bowl of internal diameter 54 cm contains a liquid. The liquid is to be filled in cylindrical bottles of radius 3 cm and height 9 cm. How many bottles are required to empty the bowl?
a)    221
b)    343
c)    8
d)    243
e)    None of these
9). How many numbers are there between 100 and 1000 such that 7 is at unit’s place?
a)    100
b)    90
c)    81
d)    82
e)    None of these
10). Find the chance of throwing a sum more than 15 in one throw with 3 dice.
a)    1/36
b)    5/216
c)    5/108
d)    11/216
e)    None of these
Answers:
1). b) 2). c) 3). a) 4). e) 5). d) 6). c) 7). a) 8). d) 9). b) 10). c)
 
Solutions:
 
1. Reqd amount = (44/0.88) = 50 quintals.
Answer: b)
 
2. Let the speed of the slower motorcyclist be x km/hr. Therefore the speed of the faster motorcyclist will be (x + 5 ) km/hr.
In 2 hrs both motorcyclists cover
2x + 2(x+5) = (4x + 10) km distance.
i.e 4x + 10 = 220 – 30
or, 4x= 180 or, x=45
Ratio = 50 : 45 = 10 : 9
Answer: c)
 
3. Let original area plouged daily = x hectares
Total area plouged = 16x
If(x + 18) hectares is plouged daily, then total area = 10(x+18)
Given that, 16x= 10(x + 18)
Or, x=30 hectares
∴ Total area of field = 16 × 30 = 480 hectares
Answer: a)
 
4. Net profit on the article
= + 20 – 15- (20×15/100) = +2%
Selling price of the article = 425 × 1.02
= Rs. 433.50
Answer: e)
 
5. By method of Alligation:
 
Ratio = 70/12 = 35.6
For each part of liquid I, 6/35 part of liquid II should be added. So for 700 ml of liquid I, 120ml of liquid II should be added.
Answer: d)
 
6. In the adjacent figure,
PQ= length of the building
RM= length of the ladder,
Given RM = 12 m, PM = 8m and
In ∆PQR
Tan 60° = [(8 + MQ) / RQ] or, RQ= [(8+MQ) / √3]
In ∆MQR
(RM)² = (MQ)² + (RQ)²
Or, (12)² = (MQ)² + [(8+MQ) / √3]2
or, MQ ≈ 8m
Height of the building = PM + MQ = 8 + 8=16m
Answer: c)
 
7. [(x³ + 3x) / (3x² + 1)] = (1241/217)
Using componendo and dividend, we get
[(x³ + 3x + 3x² + 1) / (x³ + 3x – 3x² – 1)] = [(1241+217) / (1241-217)]
or, (x+1)³ / (x-1)³ = 1458 / 1024
or, [(x+1) / (x-1)]³ =  (9/8)³
or, (x+1 / x-1) = (9/8)
or, 8x + 8 = 9x – 9 or, x=17
Answer: a)
 
8. Reqd no. of bottles = (2/3π (27)³ / π(3)²6) = 243
Answer: d)
 
9. In a three-digit no. with 7 at unit’s place, zero can’t be there at hundred’s place. So, hundred’s place can be filled with any of the digits from 1 to 9. Thus there are 9 options.
Ten’s place = digits 0 to 9 = 10 options
Unit’s place = 7 = 1 option
∴ no. of required numbers = 9 × 10 × 1 =90
Answer: b)
 
10. A throw amounting to 18 must be made up of 6, 6, 6 and this can occur in 1 way; 17 can be made up of 6, 6, 5, which can occur in 3 ways; 16 may be made up of 6, 6, 4 and 6, 5, 5, each of which arrangements can occur in 3 ways.
Therefore the no. of favorable cases is
1 + 3 + 3 + 3 = 10
Total no. of cases = 6³ = 216
∴ required probability = (10/216) = (5/108)
Answer: c)

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