SBI Clerk Mains Quantitative Aptitude Practice Set 2

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SBI Clerk Mains Quantitative Aptitude Practice Set 2

Introduction

Career in Banking is one of the most lucrative and most sought after careers. In India, Bank Recruitment Exams are primarily conducted for recruitment of Probationary Officers, Clerks & Specialist Officers. India currently[2019] has 93 commercial and 27 public sector banks out of which 19 are nationalized and 6 are SBI and its associate banks and rest two are IDBI Bank and Bharatiya Mahila Bank, which are categorized as other public sector banks. Recruitment for Bank Probationary Officers, Management Trainees, Clerks and for various other posts generally follow a 3 step recruitment process: Preliminary Exam + Mains Exam + Interview & Group Discussion. The article SBI Clerk Mains Quantitative Aptitude Practice Set 2 presents a practice set for the most sought after SBI PO recruitment. Until the year 2013, All Public Sector Banks used to conduct their own entrance test, GDs and Personal Interview for recruiting candidates. However, after 2014, IBPS started conducting recruitment Tests for 12 PSU Banks. IBPS holds a separate entrance test for recruitment.

Mains exams are very important to clear every government sector or bank related recruitment process in India. Only those candidates who are selected in the Mains round are allowed to move further up in the recruitment process. The marks obtained in the Mains exams are considered for the final merit list. Mains exams usually consist of 4 sections, with 155 questions with a time duration of 3 hours. Mains exams most certainly have negative marking.

Quiz

1. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

A. 3.6
B. 7.2
C. 8.4
D. 10

Explanation:
Speed = [$$\frac{600}{5× 60}$$ ]m/sec
= 2 m/sec.
Converting m/sec to km/hr
= 2 x$$\frac{18}{5}$$ km/hr
= 7.2 km/hr.

2. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hour, it must travel at a speed of:

A. 300 kmph
B. 360 kmph
C. 600 kmph
D. 720 kmph

Explanation:

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = $$\frac{1200}{5/3}$$ km/hr.
Required speed = 1200 x $$\frac{3}{5}$$ km/hr= 720 km/hr

3. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:

A. 50 km
B. 56 km
C. 70 km
D. 80 km

Explanation:
Let the actual distance travelled be x km.

Then, $$\frac{×}{10}$$ = $$\frac{× + 20 }{14}$$
⇒14x = 10x + 200

⇒4x = 200

⇒ x = 50 km.

4. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

A. 100 kmph
B. 110 kmph
C. 120 kmph
D. 130 kmph

Explanation:
Let speed of the car be x kmph
Then, speed of the train = $$\frac{150}{100}$$ x = $$\frac{3}{2}$$
∴ $$\frac{75}{×}$$ – $$\frac{75}{3/2}$$ x = $$\frac{125}{10 × 60 }$$
⇒$$\frac{75}{×}$$ – $$\frac{50}{×}$$ = $$\frac{5}{24}$$
⇒x = $$\frac{25 × 24}{5}$$ = 120 kmph

5. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

A. 9
B. 10
C. 12
D. 20

Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = [$$\frac{9}{54}$$ x 60] min = 10 min

1. The seven digit number 43567X is divisible by 3, where X is a single digit whole number. Find X.

A. 2
B. 5
C. 8
D. All of these

Explanation:
A number is divisible by 3 when sum of its digits is divisible by 3.
Here, sum of digits = 4 + 3 + 5 + 6 + 7 + X = 25 + X. So, X can be 2, 5, 8 which gives the sum 27, 30 and 33 respectively.
Therefore, X has 3 values here, for which the number is divisible by 3. So, the answer is option D.

2. Simplify the expression using BODMAS rule ($$\frac{3}{7}$$) of ($$\frac{4}{5}$$) of 20 ($${25}^{2}$$ – $${24}^{2}$$)

A. 336
B. 168
C. 84
D. None of these

Explanation:
($$\frac{3}{7}$$) × ($$\frac{4}{5}$$) of 20 (625 – 576)⇒ ($$\frac{3}{7}$$) × ($$\frac{4}{5}$$) × 20 × 49 =336

3. Simplify the expression using BODMAS rule (105 + 206) – 550 ÷ $${5}^{2}$$ + 10

A. 399
B. 289
C. 298
D. 299

Explanation:
(105 + 206) – 550 ÷ 52 + 10
= 311 – 550 ÷ 25 + 10
= 311 – 22 + 10
= 289 + 10 = 299

4. Simplify the expression using BODMAS rule:: ($$\frac{3}{3}$$) of ($$\frac{4}{7}$$) {(10 × 3) – (8 × 2)}

A. 6
B. 12
C. 18
D. 14

Explanation:
Applying BODMAS rule = ($$\frac{3}{2}$$) of ($$\frac{4}{7}$$) {30 – 16} = ($$\frac{12}{14}$$) × 14 = 12

5. The six-digit number 54321A is divisible by 9 where A is a single digit whole number. Find A.

A. 0
B. 2
C. 4
D. 3

Explanation:
A number is divisible by 3 when sum of its digits is divisible by 3.
Here, sum of digits = 4 + 3 + 5 + 6 + 7 + X = 25 + X. So, X can be 2, 5, 8 which gives the sum 27, 30 and 33 respectively.
Therefore, X has 3 values here, for which the number is divisible by 3.

1. A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :

A. 4 days
B. 6 days
C. 8 days
D. 12 days

Explanation:
(A + B + C)’s 1 day’s work = $$\frac{1}{6}$$
(A + B)’s 1 day’s work = $$\frac{1}{8}$$
(B + C)’s 1 day’s work = $$\frac{1}{12}$$
∴(A + C)’s 1 day’s work = [2$$\frac{1}{6}$$] – [$$\frac{1}{8}$$ + $$\frac{1}{12}$$]
= $$\frac{1}{3}$$ – $$\frac{5}{24}$$
= $$\frac{3}{24}$$
= $$\frac{1}{8}$$
So, A and C together will do the work in 8 days.

2. A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:

A. 5 days
B. 6 days
C. 10 days
D. 10$$\frac{1}{2}$$ days

Explanation:
(B + C)’s 1 day’s work = [$$\frac{1}{9}$$ + $$\frac{1}{12}$$] = $$\frac{7}{36}$$
Work done by B and C in 3 days =[ $$\frac{7}{36}$$ x 3] = $$\frac{7}{12}$$
Remaining work = [1 – $$\frac{7}{12}$$] = $$\frac{5}{12}$$
Now, $$\frac{1}{24}$$ work is done by A in 1 day
So, $$\frac{5}{12}$$ work is done by A in [24 x $$\frac{5}{12}$$] = 10 days

3. X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?

A. 13$$\frac{1}{3}$$ days
B. 15 days
C. 20 days
D. 26 days

Explanation:
Work done by X in 8 days = [$$\frac{1}{40}$$ x 8] = $$\frac{1}{5}$$
Remaining work = [1 – $$\frac{1}{5}$$] = $$\frac{4}{5}$$
Now,$$\frac{4}{5}$$ work is done by Y in 16 days.
Whole work will be done by Y in [16 x$$\frac{5}{4}$$] = 20 days
∴X’s 1 day’s work = $$\frac{1}{40}$$, Y’s 1 day’s work = $$\frac{1}{20}$$
(X + Y)’s 1 day’s work = [$$\frac{1}{40}$$ + $$\frac{1}{20}$$] = $$\frac{3}{40}$$ = $$\frac{3}{40}$$
Hence, X and Y will together complete the work in $$\frac{40}{3}$$ = 13$$\frac{1}{3}$$

4. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

A. 15
B. 16
C. 18
D. 25

Explanation:
Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).
∴[$$\frac{20 – ×}{×}$$ x 100 = 25]
⇒2000 – 100x = 25x

⇒125x = 2000

⇒ x = 16.

5. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

A. 30%
B. 70%
C. 100%
D. 250%

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.
∴Required percentage = [$$\frac{295}{420}$$ x 100]% = $$\frac{1475}{21}$$% = 70%

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