Important for :

1

C

The ratio in the initial mixture and the remaining mixture remains same.The ratio in the initial mixture = 24 : 6 = 4 : 1

Since, 18 litres is removed, 12 litres remains.

Capacity of petrol in the remaining mixture

= 12 × | 4 | = 9.6 litres |

5 |

Capacity of Diesel in the remaining mixture

= 12 × | 1 | = 2.4 litres |

5 |

Hence, option C is correct.

2

C

Priyanka and Sonam can do 1/y and 1/(y + 3)In 5 days, they can do 5 {1/y + 1/(y+3)} of the total project.

Total profit was Rs. 2400, Anchal is getting Rs. 600 as her share for the part of the project she did.

x = | 600 | = | 1 |

2400 | 4 |

So, Anchal did 1/4

→ 5 | ( | 1 | + | 1 | ) | = | 3 |

y | y + 3 | 4 |

Solving this we get y = 12

Hence, option C is correct.

3

B

14.28% = | 1 |

7 |

11.11% = | 1 |

9 |

LCM {7, 9} = 63

Let the money he had be Rs. 63x

He spend 1/7 on tickets = | 63x | = Rs. 9x |

7 |

He is left with = (63 – 9)x = Rs. 54x

Now, he spend 1/9 of 54x on snacks i.e., = | 54x | = Rs. 6x |

9 |

Remaining money = (54 – 6)x = Rs. 48x

48x is divided in the ratio 5 : 3 for lunch and transport,

For transport = | 48x × 3 | = 18x |

8 |

According to the question,

18x corresponds to 330

x correspond to = | 330 | = | 55 |

18 | 3 |

Money spend on tickets, 9x will correspond to

= | 9 × 55 | = Rs. 165 |

3 |

Hence, option B is correct.

4

D

Let the present age of father, son and daughter be F, S and D respectively.

According to the question,

5 years ago, the age of father is 2.25 times the age of son.

F – 5 = 2.25 (S – 5)

F = 2.25S – 5 (2.25 – 1)

F = 2.25S – 6.25 ...(1)

2 years hence, the age of father becomes 2.6 times the age of daughter,

F + 2 = 2.6 (D + 2)

F = 2.6D + 2 (2.6 – 1)

F = 2.6D + 3.2 ...(2)

S – D = 7 ...(3)

From (1) and (2),

2.25S – 6.25 = 2.6D + 3.2

2.25S – 2.6D = 9.45 ...(4)

Multiply (3) by 2.25 and subtract it from (4)

– 0.35D = – 6.3

D = 18 years, S = 18 + 7 = 25 years

Substituting value of D in (2),

F = 2.6 × 18 + 3.2

F = 50 years

Hence, option D is correct.

5

E

Taps T, U and V can fill in z, (z + 3) and (z + 8) hours respectively.

Capacity of tank = LCM {z, z + 3, z + 8}

In 1^{st} hour, T is opened, in 2^{nd} hour U is opened and in the 3^{rd} hour V is opened. In 3 hours, the taps were opened once, alternatively.

Capacity of tank = LCM {z, z + 3, z + 8}

In 1

According to the question, total time taken to the tank completely = 15 hours

→ | ( | 1 | + | 1 | + | 1 | ) | ||

z | z + 3 | z + 8 | = | 1 | |||||

3 | 15 |

→ | ( | 1 | + | 1 | + | 1 | ) | = | 1 |

z | z + 3 | z + 8 | 5 |

Substituting the values from the options,

→ z = 15,

→ | ( | 1 | + | 1 | + | 1 | ) | = | 138 + 115 + 90 |

15 | 18 | 23 | 2070 |

→ | 343 | ≠ | 1 |

2070 | 5 |

→ z = 20

→ | ( | 1 | + | 1 | + | 1 | ) | ≠ | 1 |

20 | 23 | 28 | 5 |

→ z = 10

→ | ( | 1 | + | 1 | + | 1 | ) | ≠ | 1 |

10 | 13 | 18 | 5 |

→ z = 18

→ | ( | 1 | + | 1 | + | 1 | ) | ≠ | 1 |

18 | 21 | 26 | 5 |

→ z = 12

→ | ( | 1 | + | 1 | + | 1 | ) | = | 5 + 4 + 3 |

12 | 15 | 20 | 60 |

→ | 12 | = | 1 |

60 | 5 |

→ | 1 | = | 1 |

5 | 5 |

Hence, option E is correct.

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Here are some important examples regarding

15

9

12

18

21

Priyanka and Sonam can do 1/y and 1/(y + 3)

In 5 days, they can do 5 {1/y + 1/(y+3)} of the total project.

Total profit was Rs. 2400, Anchal is getting Rs. 600 as her share for the part of the project she did.

x = | 600 | = | 1 |

2400 | 4 |

So, Anchal did 1/4

→ 5 | ( | 1 | + | 1 | ) | = | 3 |

y | y + 3 | 4 |

Solving this we get y = 12

Hence, option C is correct.

.

15

20

10

18

12

Taps T, U and V can fill in z, (z + 3) and (z + 8) hours respectively.

Capacity of tank = LCM {z, z + 3, z + 8}

In 1^{st} hour, T is opened, in 2^{nd} hour U is opened and in the 3^{rd} hour V is opened. In 3 hours, the taps were opened once, alternatively.

Capacity of tank = LCM {z, z + 3, z + 8}

In 1

According to the question, total time taken to the tank completely = 15 hours

→ | ( | 1 | + | 1 | + | 1 | ) | ||

z | z + 3 | z + 8 | = | 1 | |||||

3 | 15 |

→ | ( | 1 | + | 1 | + | 1 | ) | = | 1 |

z | z + 3 | z + 8 | 5 |

Substituting the values from the options,

→ z = 15,

→ | ( | 1 | + | 1 | + | 1 | ) | = | 138 + 115 + 90 |

15 | 18 | 23 | 2070 |

→ | 343 | ≠ | 1 |

2070 | 5 |

→ z = 20

→ | ( | 1 | + | 1 | + | 1 | ) | ≠ | 1 |

20 | 23 | 28 | 5 |

→ z = 10

→ | ( | 1 | + | 1 | + | 1 | ) | ≠ | 1 |

10 | 13 | 18 | 5 |

→ z = 18

→ | ( | 1 | + | 1 | + | 1 | ) | ≠ | 1 |

18 | 21 | 26 | 5 |

→ z = 12

→ | ( | 1 | + | 1 | + | 1 | ) | = | 5 + 4 + 3 |

12 | 15 | 20 | 60 |

→ | 12 | = | 1 |

60 | 5 |

→ | 1 | = | 1 |

5 | 5 |

Hence, option E is correct.

1 hour

3 hours

2 hours

4 hours

2.5 hours

Let the distance travelled with speed 50 kmph be a and the distance travelled with 100 kmph be (190 – a) km respectively.

According to the question,

190 – a | + | a | = 3 |

100 | 50 |

190 – a + 2a | = 3 |

100 |

a = 110 km

Now speed is 55 kmph,

Time taken = | 110 | = 2 hours |

55 |

Hence, option C is correct.

All the above questions through, you must make sure about arithmetic miscellaneous questions which have been seen in the last few years. We hope this above *Quantitative Aptitude for SBI Clerk 2020* will clear your all doubts regarding bank clerk exams.

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Team Smartkeeda