## 1 Jun 2016

### SBI Clerk 2016- Practice Aptitude Questions (Arithmetic Quiz)

SBI Clerk 2016- Practice Aptitude Questions (Arithmetic Quiz) Set-61:
Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

1). A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm . What is the area of the circle?
a)   88 cm2
b)   1250 cm2
c)   616 cm2
d)   Cannot be determined
e)   None of these

2). A alone can complete a piece of work in 8 days. Work done by ‘B’ alone in one day is half of the work done by ‘A’ alone in one day. In how many days can the work be completed if A and B work together?
a)   6 (1/3)
b)   5 (2/3)
c)   5 (1/3)
d)   6 (2/3)
e)   None of these

3). Mr Shirke distributes the money he has among his 2 sons, 1 daughter and wife in such a way that each son gets double the amount of the daughter and the wife gets double the amount of each son. If each son gets Rs. 8,500, what was the total amount distributed?
a)   Rs. 38,250
b)   Rs. 34, 250
c)   Rs. 38,500
d)   Rs. 32,500
e)   None of these

4). The difference between a two-digit number and the number obtained by interchanging the digits is 18. The sum of the digits is 10 and the digit at the ten’s place is bigger than the digit at the unit’s place. What is the two-digit number?
a)   82
b)   46
c)   64
d)   73
e)   None of these

5). Sarita started a business investing Rs. 50,000. After six months Abhishek joined her with Rs. 75,000. After another six months Nisha also joined them with Rs. 1.25 lakh. Profit earned at the end of 2 years from when Sarita started the business should be distributed among Sarita, Abhishek and Nisha in what respective ratio?
a)   4 : 5 : 6
b)   8 : 9 : 10
c)   8 : 9 : 12
d)   4 : 5 : 8
e)   None of these

6). Srinivas sold as article for Rs 460 and earned a profit of 15%. At what price should it have been sold so as to earn a profit of 20%?
a)   Rs. 483
b)   Rs. 480
c)   Rs. 498
d)   Rs. 485
e)   None of these

7). Average weight of 10 boys is more than the average weight of 15 girls by 5 kg. If the total weight of the 10 boys is 550kg, what is the average weight of the 10 boys and 15 girls together?
a)   52 kg
b)   52.5 kg
c)   53 kg
d)   53.5 kg
e)   None of these

8). If the numerator of a fraction is increased by 10% and the denominator is increased by 20%, the fraction thus obtained is 22/45. What was the original fraction?
a)   18/15
b)   11/15
c)   8/15
d)   Cannot be determined
e)   None of these

9). Three-fifths of a number is equal to 85% of another number. What is the ratio of the first number to the second?
a)   12 : 7
b)   12 : 17
c)   7 : 12
d)   17 : 12
e)   None of these

10). A car starts running at the speed of 40 km per hour. The speed of the car increases by 2 km at the end of every one hour. What will be the distance covered at the end to ten hours from the start of the journey?
a)   590 km
b)   480 km
c)   490 km
d)   520 km
e)   None of these

1). c) 2). c) 3). a) 4). c) 5). b) 6). b) 7). a) 8). c) 9). d)  10). c)

Solutions:

1). Perimeter of a rectangle = 2 (length + breath)
2 (26 + 18) = 88 cm2
Now, since perimeter of a circle = 2πr
Therefore 2πr = 88; r = 44 / π
Area of circle = πr2  = π × (44 / π ) × (44 / π) = (44 × 44) / (22 / 7)
= 616 cm2.

2). A does the work in 8 days
B does the work in 16 days
:. A+B do the work in (8 × 16) /(8 + 16)
= 16 /3  = 5 (1/3) days
Quicker Approach: B is half efficient as A
So A + B = 1.5 A = (3/2)A
:. Required no. of days = 8 × (2/3) = 16/3=5 (1/3)days

3). 2 sons get= 8500 × 2 = Rs.17000
1 daughter gets = 8500 ÷ 2=Rs.4250
Wife gets = 8500 × 2 = Rs.17000
:. Total = 38250

4). (10x+y)-(10y+x)= 18
à 9(x-y)=18
:. x-y=2
Also, given that x+y=10
:. x=6 and y=4
:. The number is 64.

5). Sarita : Abhishek : Nisha
= 50,000 × 24 : 75,000 × 18 : 1,25,000×12
= 2 × 24 : 3 × 18 : 5 × 12
= 2 × 4 : 3 × 3 : 5 × 2
=8 : 9 : 10

6). Cost price = 460(100/115) = Rs.400
Required selling price = 400 (120/100) = Rs.480

7). Average wt of 10 boys = (550/10) = 55 kg
Average wt of 15 girls= 55-5 = 50kg
:. Average wt of 10 boys and 15 girls
= (10×15+15×50)/25
=52 kg

8). Let the fraction be x/y. Then
(1.1x) / (1.2y) = (22 / 45)
:. (x/y) =  (22×1.2) / (45×1.1)
= (22×12) / (45×11) = (8/15)
Note: Here some one may say that (8/15) (x/y)
= (16/30), (24/45), (32/60)….. are other such fractions;
So our answer should be “cannot be determined”. But it is not true to say so. We treat all of them as the same fraction.

9). (3/5)x = 85% of y
:. (x/y) = (85/100) × (5/3)
= (17/12) = 17 : 12

10). Distance covered at the end of 10th hour
= 40 + 42 + 44 +……+ 58
=(40×10) +(2+4+…..+18)
=400+2 (1+2+……49)
=400 + [2 (9×10) / 2]= 490km
By Direct Formula: As the distances are in Arithmetic Progression,
Sum= n/2 [2a+(n-1)d]
= 10/2 [2×40+(10-1)×2]
=5[80+18]= 490 km