Saturday, 22 January 2022

SBI Clerk 2022 Prelims Day 1 : Numerical ability Online Test 1

SBI Clerk 2022 Prelims numerical ability Online Test Practice free .. start prepare now . hops SBI Clerk Notification get released Soon. 


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1.  Directions(Q.1 to Q.7 ):


A mixture of 30 litres contains alcohol and water in ratio 3 : 7. How much alcohol must be added to this mixture so that the ratio of alcohol and water becomes 2 : 3 ?

a
  4 ltr.
b
 6ltr.
c
 7 ltr.
d
 5 ltr.
e
 None of these
Answer & Explanation
Answer : Option D
Explanation :

Let the alcohol added be x liter

ATQ

9+x30+x=25\frac{9+x}{30+x}=\frac{2}{5}

Or, 45 + 5x = 60 + 2x

Or, 3x = 15

Or, x = 5 ltr.

2. 

A shopkeeper buys rice from wholesaler at 30% discount on marked price. If shopkeeper gives discount of 20% on marked price while selling the same rice, find his net profit/loss percentage.

a
  1212%12 \frac{1}{2} \%
b
 2127%21 \frac{2}{7} \%
c
 1427%14 \frac{2}{7} \%
d
 1623%16 \frac{2}{3} \%
e
 2327%23 \frac{2}{7} \%
Answer & Explanation
Answer : Option C
Explanation :

Let, he buys 1000 gm rice at Rs. 1000 i.e.

Rs. 1/gm.

CP for shopkeeper =70100×1000==\frac{70}{100} \times 1000= Rs. 700

SP for shopkeeper =80100×1000==\frac{80}{100} \times 1000= Rs. 800

Profit %=(800700)700×100=1427%\%=\frac{(800-700)}{700} \times 100=14 \frac{2}{7} \%

3. 

If ratio of length, breadth and height of a cuboid is 1 : 2 : 3 and its area is 88 cm², then find the volume of a cube having edge length equal to the breadth of the cuboid.

a
  44 cm³
b
 74 cm³
c
 68 cm³
d
 64 cm³
e
 None of these
Answer & Explanation
Answer : Option D
Explanation :

Let, length, breadth and height be x, 2x and

3x respectively

Then,

2(x×2x + 2x×3x + x×3x) = 88

or, 2(2x² + 6x² + 3x²) = 88

or, 22x² = 88

or, x² = 4 ⇒ x = 2

Volume of cube = (2x)³ = (4)³ = 64 cm³

4. 

The probability of Ram speaking truth is ⅓ and that of Shyam is ⅖. A question is asked from both of them. What is the probability that one

a
  615\frac{6}{15}
b
 715\frac{7}{15}
c
 815\frac{8}{15}
d
 515\frac{5}{15}
e
 None of these
Answer & Explanation
Answer : Option B
Explanation :

Required probability =13×35+23×25=715=\frac{1}{3} \times \frac{3}{5}+\frac{2}{3} \times \frac{2}{5}=\frac{7}{15}

5. 

The length of train A is twice that of train B and speed of train A is half of that of train B. If train A crosses a man in 4 sec. then find how long will train B take to cross train A if they go in same direction.

a
  3s
b
 6s
c
 5s
d
 4s
e
 None of these
Answer & Explanation
Answer : Option B
Explanation :

Let, length of train A be 2x m and speed be y m/s

Then, length of train B is x m and speed of

train B is 2y m/s.

ATQ

2xy=4\frac{2 x}{y}=4

Required time =(2x+x)(2yy)=3xy=3×2yy=6s=\frac{(2 x+x)}{(2 y-y)}=\frac{3 x}{y}=\frac{3 \times 2 y}{y}=6 s.

6.  Directions(Q.6 to Q.10 ):

What will come in place of ’x’ in the following questions. Find the approximate value.


17.993×13.98=x\frac{17.99}{3} \times \frac{1}{\sqrt{3.98}}=\sqrt{x}

a
  2
b
 4
c
 6
d
 3
e
 9
Answer & Explanation
Answer : Option E
Explanation :

183×14=x\frac{18}{3} \times \frac{1}{\sqrt{4}}=\sqrt{x}

or, 3 = √x or, x = 9

7. 

13\frac{1}{3} of 899.8÷2199.9=x899.8 \div \frac{2}{199.9}=\mathrm{x}

a
  43000
b
 24000
c
 20000
d
 30000
e
 25000
Answer & Explanation
Answer : Option D
Explanation :

13×900×2002=x\frac{1}{3} \times 900 \times \frac{200}{2}=\mathrm{x} or, 30000 = x

8. 

22.9 + 13.1 – 9.99% of 200.22 = x

a
  16
b
 12
c
 17
d
 13
e
 24
Answer & Explanation
Answer : Option A
Explanation :

23+1310100×200=x23+13-\frac{10}{100} \times 200=\mathrm{x}

or, x = 36 – 20 = 16

9. 

19.90%19.90 \% of 160×4.0163.99=x160 \times \frac{4.01}{63.99}=\mathrm{x}

a
  2
b
 4
c
 3
d
 6
e
 8
Answer & Explanation
Answer : Option A
Explanation :

20100×160×464=x\frac{20}{100} \times 160 \times \frac{4}{64}=\mathrm{x} \quad or, x=2\mathrm{x}=2

10. 

67.8915.11×89.917=19.99+x\frac{67.89}{15.11} \times \frac{89.9}{17}=19.99+\mathrm{x}

a
  8
b
 4
c
 6
d
 5
e
 2
Answer & Explanation
Answer : Option B
Explanation :

6815×9017=20+x\frac{68}{15} \times \frac{90}{17}=20+\mathrm{x} or, x = 4

11.  Directions(Q.11 to Q.15 ):

Study the table carefully and answer the questions. Table given below shows the number of items sold by five persons on five different days

Note: Some data are missing, calculate the missing data if required.


If average of items sold by D on Monday and Tuesday is 245. Then items sold by D on Monday is what percent of item sold by E on Friday ?

a
  50%
b
 None of these
c
 60%
d
 70%
e
 75%
Answer & Explanation
Answer : Option D
Explanation :

Items sold by D on Monday = 245 × 2 – 210

= 280

Required %=280400×100=70%\%=\frac{280}{400} \times 100=70 \%

12. 

If items sold by C on Thursday is average of items sold by C on Wednesday and Friday then find total items sold by C ?

a
  None of these
b
 1900
c
 1700
d
 1750
e
 1850
Answer & Explanation
Answer : Option B
Explanation :

Required total =(410+3502)+250+520+=\left(\frac{410+350}{2}\right)+250+520+

410+350=1900410+350=1900

13. 

If total item sold by A and B including all fivedays is 2000 and 2200 respectively. Then item sold by B on Tuesday is what percent more/less than item sold by A on Friday ?

a
  42.5%
b
 23.5%
c
 37.5%
d
 32.25%
e
 47%
Answer & Explanation
Answer : Option C
Explanation :

Items sold by A on Friday = 2000 – (420 +360 + 280 + 540)

= 2000 – 1600 = 400

Items sold by B on Tuesday

= 2200 – (440 + 240 + 510 +460) = 550

Required %=(550400)400×100=37.5%\%=\frac{(550-400)}{400} \times 100=37.5 \%

14. 

If ratio of items sold by B and C together on Thursday to items sold by C and E together on same day is 2 : 1. Then find item sold by C on Thursday ?

a
  150
b
 190
c
 210
d
 None of these
e
 170
Answer & Explanation
Answer : Option B
Explanation :

Let items sold by C on Thursday be x.

ATQ,

510+xx+160=21\frac{510+x}{x+160}=\frac{2}{1}

510 + x = 2x + 320

x = 190

15. 

What is difference of item sold by A on Monday and Friday. If total item sold by A is 1800 ?

a
  225
b
 210
c
 None of these
d
 320
e
 220
Answer & Explanation
Answer : Option E
Explanation :

Item sold by A on Friday = 1800 – 1600= 200

Required difference = 420 – 200 = 220

16.  Directions(Q.16 to Q.25 ):

What will come in place of ’x’ in the following questions. Find the exact value.


2426÷34+x=31\frac{242}{6} \div \frac{3}{4}+x=31

a
  2159\frac{-215}{9}
b
 2259\frac{-225}{9}
c
 2359\frac{-235}{9}
d
 2059\frac{-205}{9}
e
 None of these
Answer & Explanation
Answer : Option D
Explanation :

2426×43+x=31\frac{242}{6} \times \frac{4}{3}+\mathrm{x}=31

or, 242×29+x=31\frac{242 \times 2}{9}+\mathrm{x}=31

or, x=314849x=31-\frac{484}{9}

or, x=2794849=2059x=\frac{279-484}{9}=\frac{-205}{9}

17. 

4349.1 + 2256.4 – 1151.2 – 1244.3 = x

a
  4321
b
 4230
c
 4210
d
 4530
e
 None of these
Answer & Explanation
Answer : Option C
Explanation :

6605.5 – 2395.5 = x

or, x = 4210

18. 

729×19224÷3%729 \times \frac{1}{9^{2}}-24 \div 3 \% of 10=x10=\mathrm{x}

a
  -71
b
 -70
c
 -72
d
 -76
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :

729×18124×1003×10=x729 \times \frac{1}{81}-24 \times \frac{100}{3 \times 10}=\mathrm{x}

or, 9 – 80 = x

or, x = –71

19. 

1728 ÷ 12³ – 11³ = x

a
  -1330
b
 -1142
c
 -1342
d
 -1332
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :

1728×1123113=x1728 \times \frac{1}{12^{3}}-11^{3}=x

or, 1728×117281331=x1728 \times \frac{1}{1728}-1331=\mathrm{x}

or, x = –1330

20. 

1089 × 101 + 36 × 44 = x

a
  111573
b
 111576
c
 111546
d
 111453
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :

1089 × (100 + 1) + (40 – 4) (40 + 4) = x

or, 1089 + 108900 + 40² – 4² = x

or, 109989 + 1600 – 16 = x

or, 111573 = x

21. 

1001 × 1120 – 1120 × 999 + 55 = x

a
  2295
b
 2234
c
 2245
d
 2278
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :

1120 (1001 – 999) + 55 = x

or, 1120 × 2 + 55 = x

or, 2240 + 55 = x

or, x = 229

22. 

27\frac{2}{7} of 343+512256=x343+\frac{512}{256}=\mathrm{x}

a
  100
b
 90
c
 80
d
 110
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :

27×343+2=x\frac{2}{7} \times 343+2=x

or, 98 + 2 = x

or, x = 100

23. 

62 ÷ 31% of 200 + x = 4

a
  2
b
 4
c
 3
d
 5
e
 None of these
Answer & Explanation
Answer : Option C
Explanation :

62×10031×200+x=462 \times \frac{100}{31 \times 200}+x=4

or, 1 + x = 4 or, x = 3

24. 

34324×53+168=x\frac{3432}{4} \times \frac{5}{3}+168=\mathrm{x}

a
  1546
b
 1432
c
 1598
d
 1234
e
 None of these
Answer & Explanation
Answer : Option C
Explanation :

34324×53+168=x\frac{3432}{4} \times \frac{5}{3}+168=\mathrm{x}

or, 858×53+168=x858 \times \frac{5}{3}+168=x

or, 286 × 5 + 168 = x

or, 1430 + 168 = x

or, x = 1598

25. 

64×198+3408×417=?64 \times 198+\frac{340}{8} \times \frac{4}{17}=?

a
  12345
b
 12435
c
 12682
d
 14567
e
 None of these
Answer & Explanation
Answer : Option C
Explanation :

64×(2002)+202=x64 \times(200-2)+\frac{20}{2}=x

or, 12800 – 128 + 10 = x

or, x = 12682

26.  Directions(Q.26 to Q.30 ):

Find the value of question mark (?) in following number series:


10, 13, 18, 26, 38, ?

a
  43
b
 55
c
 45
d
 65
e
 None of these
Answer & Explanation
Answer : Option B
Explanation :
27. 

2, 7, 25, 105, 531, ?

a
  3567
b
 3241
c
 2341
d
 3193
e
 None of these
Answer & Explanation
Answer : Option D
Explanation :
28. 

3, 4, 8, 17, 33, ?

a
  58
b
 45
c
 54
d
 47
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :
29. 

7, 7, 10, 18, 33, ?

a
  57
b
 54
c
 65
d
 34
e
 None of these
Answer & Explanation
Answer : Option A
Explanation :
30. 

5, 7, 13, 25, 45, ?

a
  45
b
 75
c
 65
d
 86
e
 None of these
Answer & Explanation
Answer : Option B
Explanation :
31.  Directions(Q.31 to Q.35 ):

Study the passage carefully and answer the questions based on it.

Marks obtained by X in Physics is 160 which is 40% of total marks obtained by him. And marks obtained by Z in chemistry is ¾th of marks obtained by X in Physics. Ratio of total marks obtained by X to Z is 4 : 3. Marks obtained by Y in Maths is equal to marks obtained by Z in Chemistry. And total marks obtained by Y in Physics and Chemistry together is half of total marks obtained by X.


If ratio of marks obtained by Z in Physics and Maths is 2 : 1, then find average marks obtained by X and Z in Physics ?

a
  None of these
b
 150
c
 160
d
 120
e
 140
Answer & Explanation
Answer : Option E
Explanation :
Marks obtained X in Physics = 160 Total marks obtained by X=16040×100=400\mathrm{X}=\frac{160}{40} \times 100=400 Marks obtained by X in Chemistry & Maths together = 400 – 160 = 240 Total marks obtained by Z=4004×3=300\mathrm{Z}=\frac{400}{4} \times 3=300 Marks obtained by Z\mathrm{Z} in Chemistry =34×160=120=\frac{3}{4} \times 160=120 Marks obtained by Z in Physics and Maths = 300 – 120 = 180 Marks obtained by Y in Maths = 120 Marks obtained by Y in Physics and Chemistry Together =4002=200=\frac{400}{2}=200 Total marks obtained by Y = 200 + 120 = 320

Marks obtained by Z in Physics

= 180 × ⅔ = 120

Required average =120+1602=140=\frac{120+160}{2}=140

32. 

What is ratio of marks obtained by Z in Physics and Maths together to marks obtained by X in Chemistry and Physics together ?

a
  2 : 5
b
 4 : 3
c
 cannot be determined
d
 1 : 2
e
 2 : 3
Answer & Explanation
Answer : Option C
Explanation :
Marks obtained X in Physics = 160 Total marks obtained by X=16040×100=400\mathrm{X}=\frac{160}{40} \times 100=400 Marks obtained by X in Chemistry & Maths together = 400 – 160 = 240 Total marks obtained by Z=4004×3=300\mathrm{Z}=\frac{400}{4} \times 3=300 Marks obtained by Z\mathrm{Z} in Chemistry =34×160=120=\frac{3}{4} \times 160=120 Marks obtained by Z in Physics and Maths = 300 – 120 = 180 Marks obtained by Y in Maths = 120 Marks obtained by Y in Physics and Chemistry Together =4002=200=\frac{400}{2}=200 Total marks obtained by Y = 200 + 120 = 320

Cannot be determined

33. 

If marks obtained by X in Maths is 50% more than marks obtained by him in Chemistry. Then find difference of marks obtained by X in Maths and marks obtained by Y in Maths ?

a
  14
b
 32
c
 24
d
 12
e
 28
Answer & Explanation
Answer : Option C
Explanation :
Marks obtained X in Physics = 160 Total marks obtained by X=16040×100=400\mathrm{X}=\frac{160}{40} \times 100=400 Marks obtained by X in Chemistry & Maths together = 400 – 160 = 240 Total marks obtained by Z=4004×3=300\mathrm{Z}=\frac{400}{4} \times 3=300 Marks obtained by Z\mathrm{Z} in Chemistry =34×160=120=\frac{3}{4} \times 160=120 Marks obtained by Z in Physics and Maths = 300 – 120 = 180 Marks obtained by Y in Maths = 120 Marks obtained by Y in Physics and Chemistry Together =4002=200=\frac{400}{2}=200 Total marks obtained by Y = 200 + 120 = 320

Let marks obtained by X in Chemistry be a.

ATQ,

a+150100a=240a+\frac{150}{100} a=240

a = 96

Required difference = 96 × 1.5 – 120

= 144 – 120 = 24

34. 

If Max. marks in each subject is 200. Then find percent of marks obtained by Y ?

a
  42 ⅔%
b
 61 ⅓%
c
 None of these
d
 53 ⅓%
e
 47 ⅔%
Answer & Explanation
Answer : Option D
Explanation :
Marks obtained X in Physics = 160 Total marks obtained by X=16040×100=400\mathrm{X}=\frac{160}{40} \times 100=400 Marks obtained by X in Chemistry & Maths together = 400 – 160 = 240 Total marks obtained by Z=4004×3=300\mathrm{Z}=\frac{400}{4} \times 3=300 Marks obtained by Z\mathrm{Z} in Chemistry =34×160=120=\frac{3}{4} \times 160=120 Marks obtained by Z in Physics and Maths = 300 – 120 = 180 Marks obtained by Y in Maths = 120 Marks obtained by Y in Physics and Chemistry Together =4002=200=\frac{400}{2}=200 Total marks obtained by Y = 200 + 120 = 320

Required %=320600×100=531/3%\%=\frac{320}{600} \times 100=531 / 3 \%

35. 

Marks obtained by Y in Physics and Chemistry together is what percent of total marks obtained by Z ?

a
  None of these
b
 50%
c
 33 ⅓%
d
 66 ⅔%
e
 75%
Answer & Explanation
Answer : Option D
Explanation :
Marks obtained X in Physics = 160 Total marks obtained by X=16040×100=400\mathrm{X}=\frac{160}{40} \times 100=400 Marks obtained by X in Chemistry & Maths together = 400 – 160 = 240 Total marks obtained by Z=4004×3=300\mathrm{Z}=\frac{400}{4} \times 3=300 Marks obtained by Z\mathrm{Z} in Chemistry =34×160=120=\frac{3}{4} \times 160=120 Marks obtained by Z in Physics and Maths = 300 – 120 = 180 Marks obtained by Y in Maths = 120 Marks obtained by Y in Physics and Chemistry Together =4002=200=\frac{400}{2}=200 Total marks obtained by Y = 200 + 120 = 320

Required %=200300×100=662/3%\%=\frac{200}{300} \times 100=662 / 3 \%

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