# SBI Clerk 2018: Quantitative Aptitude Quiz for Prelims – 2

Hello and welcome to exampundit. Here is a set of Quantitative Aptitude Quiz for Prelims exam of SBI Clerk 2018.

1.Deepak and Deepika run a race between points A and B, 5 km apart. Deepak starts at 9 a.m. from A at a speed of 5km/hr, reaches B and returns to A at the same speed. Deepika starts at 9.45 a.m. from A at a speed of 10 km /hr, reaches B and comes back to A at the same speed. At what time does Deepika overtakes Deepak?

a) 10.20 a.m.

b) 10.30 a.m.

c) 10.40 a.m.

d) 10.50 a.m.

e) Cannot be determined

Option: B

Explanation:

Both Deepak and Deepika are reaching B from A and then returning . When Deepika overtakes Deepak, the distances traveled by both of them are equal. Therefore, the times taken by them are inversely proportional to their speeds.

Let Deepak take t minutes for travel, Deepika overtakes him. As Deepika started 45 minutes after Deepak he takes t-45 minutes.

=> t/ (t-45) = speed Deepika / Speed Deepak =2

=>t=90 minutes

Therefore Deepika overtakes Deepak at 9 a.m + 90 minutes = 10.30 a.m

2.Caitlin goes to the supermarket with Rs. 500 in her walet to buy strawberries. The prices of strawberries have decreased by 10 % so she could buy 2 kg more with the amount she had. What was the original price of the strawberries?

a) Rs 30

b) Rs 25

c) RS 27.77

d) Rs 32.33

e) None

Option: C

Explanation:

If the price of strawberries decreases by 10 %, Caitlin would save 10 % of her money i.e. Rs 50 to buy the same amount as before in Rs 50, she can buy 2 kg more, therefore the current price of 1 kg strawberries is Rs. 25.

This price is after 10 % deduction. Hence, original price = 25/0.9= Rs 27.77

3.Anil, Ashok and Ankit are three singers among 7 different singers, scheduled to sings in a function. What is the probability that Anil will speak before Ashok who sings before Ankit? ( it is not necessary that they sing consecutively)

a) 3

b) 2

c) 1/6

d) 5/6

e) 1/7

Option: C

Explanation:

Anil, Ashok, Ankit can be arranged in 3!= 6 ways out of which only one is favorable.

Probability = 1/6

4. Anil and Sunil have to paint 810 and 900 canvas respectively in the same period. But Anil completes his work 3 days before time and Sunil completes in 6 days before time. How many canvas did A paint per hour if B paints 21 canvas per hour?

a) 45

b) 54

c) 72

d) 100

e) None

Option: B

Explanation:

Solve through options

810/54 – 900/75= 3= (6-3=3)

15-12=3

3=3

Hence 4 days option b

5.In Tilak clubhouse, there are as many halls as there are the number of chairs in each hall and not more than one person can have the same chair. If the middlemost hall accommodating 25 persons is filled with 71.428% of its capacity, then find the maximum number of persons in the clubhouse that can be accommodated if it has minimum 20 % chairs always vacant?

a) 500

b) 786

c) 980

d) 476

e) Cannot be determined

Option: C

Explanation:

The total persons in each hall= 25x(7/5)=35

Total number of chairs= 352= 1225

Maximum available capacity= 1225X0.8=980 chairs

6. A municipality tanker for public use in Nagpur has 3 pipes connected to it. Three taps P, Q and R can fill the tanker in 12, 15 and 20 hours respectively. If P is always open and Q and R are opened for one hour each alternately, the tanker will be full in?

a) 6.5 hours

b) 7 hours

c) 11 hours

d) 9 hours

e) None

Option: B

Explanation:

(P + Q)will fill in 1 hour = (1/12 + 1/15) = 3/20 part

(P + Q)will fill in 1 hour = (1/12 + 1/20) = 2/15 part

Part filled in 2 hrs = (3/20 + 2/15) = 17/60

Part filled in 6 hrs = 3 * 17/60 = 17/20

Remaining part = 1 – 17/20 = 3/20

Now, it is the turn of P and Q and 3/20 part is filled by P and Q in 1 hour.

Total time taken to fill the tank = (6 + 1) = 7 hrs.

7. Rakesh and Ramesh run a 100 m race, where Rakesh beats Ramesh by 10 metres. To do a favour to Ramesh, Rakesh starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

a) Rakesh and Ramesh reach the finishing line simultaneously

b) Ramesh beats Rakesh by 1 metre

c) Ramesh beats Rakesh by 11 metres

d) Rakesh beats Ramesh by 1 metre

e) None of the above

Option: D

Explanation:

In first case Rakesh travels 100 m and Ramesh travels 90 m in the same time.

Therefore, the ratio of their speeds is 10:9

In the second case, Rakesh is travelling 110 m to reach the finishing line. In the same

time, the distances travelled by both of them will again be in the same ratio as their old speeds.

Distance Ramesh/Distance Rakesh = Speed Ramesh / Speed Rakesh

=>Distance Ramesh/110=9/10

=>Distance Ramesh= 99m

Therefore Ramesh has travelled only 99 m when Rakesh reached the finish

line(100m).

Hence Rakesh beats Ramesh by 1 m.

8. Abram can sow a field in 5 days. He invites Abraham and Ashley who can sow 3/4th as fast as he can to join him. He also invites Mitch and Harrison who can sow only 1/5th as fast as he can to join him. If the five persons sow the same field and start together, how long will they take to finish the complete sowing of the field?

a) 100/58 days

b) 29 days

c) 112/37 days

d) 47 days

e) None

Option: A

Explanation:

Let the total work = 100 units.

Abram’s one day work = 100/5= 20 units.

Abraham and Ashley’s 1 day work= ¾ x 20 = 15 units

Mitch and Harrison’s 1 day work= 1/5 x 20= 4 units

In one day all of them can do = 20+15+15+4+4= 58 units of work.

Hence they complete the work in 100/58 days.

9.Subham wholesale pvt ltd. , a wholesale agency cheats both while buying and selling goods. While buying goods from the farmers, they use a weighing machine that shows 1000g for 1100g . While selling, their machine shows 1100 g for 1000g. If they sell goods at cost price only, determine their profit percentage in the whole transaction.

a) 31%

b) 21%

c) 11%

d) 12%

e) 15.4%

Option: B

Explanation:

Let price of 1 g be 1 rupee

They sell 1000g as 1100g

They will sell 1100 g as 1100/1000  x 1100= 1210 g

Which means they get rs 1210

Their profit percentage= [(1210-1000)/1000]x100= 21 %

10.Aniket plants 5 trees to Sanket’s 3 but in case of Aniket only 1 plant survives in 3 plantings while in case of Sanket one plant survives in 2 plantings. When 27 plants planted by Sanket died, number of plants survived in Aniket’s plantings is

a) 30

b) 60

c) 72

d) 90

e) None

Option: A

Explanation:

Let the total number of plantings be x

plantings by Aniket= 5/8x

plantings by Sanket= 3/8x

survived plants of Aniket=1/3  of  5/8  of x=  5x/24

died plants of Sanket= ½ of 3/8  of x= 3x/16

3x/16=27=> x=144

Plants survived  by Aniket= 5x/24=30

Regards

Team EP

Average rating / 5. Vote count:

## We are sorry that this post was not useful for you!

Let us improve this post!