Important for :

1

A

Let the weight of Seema be x, weight of Monisha is (x + 5) kg, weight of Reema is (x + 15) kgAccording to the question,

[{x + 15 + x + 5}/2] | = 14 : 11 |

[{x + 5 + x}/2] |

2x + 20 | = | 14 |

2x + 5 | 11 |

11 (2x + 20) = 14 (2x + 5)

x = 25

Seema = 25 kg, Reema = 40 kg

Reqd. average weight = | 25 + 40 | = 32.5 kg |

2 |

Hence, option A is correct.

2

D

The ratio of the number of clerks and teachers is 5 : 4.Let the number of clerks and teachers are 50 and 40 respectively.

Clerks: 50 → (20%) → | 50 × 20 | = 10 |

100 |

Teachers: 40 → (25%) → | 40 × 25 | = 10 |

100 |

20 people have their own vehicles.

People who does not have their own vehicle = 90 – 20 = 70

Reqd. % = | 70 × 100 | = 77.78% |

90 |

Hence, option D is correct.

3

A

Let us suppose that sum Rs. a is invested in both the schemesThe simple interest is offered by scheme A,

a × R × 2 | = 1000 |

100 |

aR = 50000 ...(1)

The compound interest is offered by scheme B,

a | { | ( | 1 + | R | ) |
^{2} |
– 1 | } | = 1062.5..........(2) |

100 |

Dividing (1) by (2)

R | = | 50000 | |||||

( | 1 + | R | ) |
^{2} |
– 1 | 1062.5 | |

100 |

R | = | 10000 | ||||

( |
R^{2} |
+ | 2R | ) | 212.5 | |

100^{2} |
100 |

1 | = | 10000 | ||||

( | R | + | 2 | ) | 212.5 | |

100^{2} |
100 |

Solving through options,

Take R = 25%

⇒ | 1 | ||

25 | + | 2 | |

100^{2} |
100 |

⇒ | 400 | ≠ | 160 |

9 | 3 |

Take R= 30%

⇒ | 1 | ||

30 | + | 2 | |

100^{2} |
100 |

⇒ | 2500 | ≠ | 160 |

23 | 3 |

Take R= 12.5%

⇒ | 1 | ||

12.5 | + | 2 | |

100^{2} |
100 |

⇒ | 10000 | = | 10000 |

212.5 | 212.5 |

R = 12.5% satisfies the above given condition.

Required sum = 12.5% + 12.5% = 25%

Hence, option A is correct.

Rs. 1000 is the SI earned in 2 years. So the SI for one year is Rs. 500

Rs. 1062.5 is the CI earned in 2 years.

In CI , for the first year interest will be Rs. 500 and then for the second year interest will be (Rs. 500 + interest on Rs. 500 for the first year)

So the extra Rs. 62.5 is the interest earned on Rs. 500

Therefore, Rate of interest = (62.5/500)* 100 = 12.5%

Required sum = 12.5% + 12.5% = 25%

Hence, option A is correct.

4

B

The age of Sukh = z years, Geet = (z + 4) years, Veer = (z – 5) yearsAccording to the question,

The ratio of age of Geet and Veer 4 years ago is 10 : 7

z + 4 – 4 | = | 10 |

z – 5 – 4 | 7 |

z = 30 years

Sukh’s present age = 30 years

Geet’s present age = 34 years

Veer’s present age = 25 years

Reqd. average = | 30 + 34 + 25 | = 29.67 years |

3 |

Hence, option B is correct.

5

A

Let the number of toys of each variety sold be x, y and z respectively.The total cost is 200x + 150y + 50z = 600

He sold at least one of each variety.

Amount received by selling one of each toy is = 200 + 150 +50 = 400

Amount left with him = 600 – 400 = 200

For the number of toys sold to be maximum he has to sell toys of the third variety for the remaining amount 200.

For Rs. 200 he can sell | 200 | = 4 toys of this variety. |

50 |

The maximum number of toys he can sell = 1 + 1 + 5 = 7 toys

Hence, option A is correct.

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Here are some important examples regarding

A. 25.5 hours

B. 20.5 hours

C. 22.5 hours

D. 21.5 hours

E. 27.5 hours

Let the speed of boat be x km/hr

Speed of the current = 4 km/hr

According to the question,

60 | + | 60 | = 20 |

x + 4 | x – 4 |

x = 8 km/hr

Now the boat goes 30 kms downstream,

Time taken to move 30 kms downstream

= | 30 | = 2.5 hours |

8 + 4 |

Total time taken = 20 + 2.5 = 22.5 hours

Hence, option C is correct.

.

A. Rs . 15000

B. Rs. 14500

C. Rs. 12500

D. Rs. 12600

E. Rs. 14000

Cost price of Trouser is 15% more than cost price of a shirt.

Let the cost price of shirt be Rs. 100x and cost price of Trouser is Rs. 115x.

Marked price of Trouser is 20% above its cost price

MRP of Trouser = (115x) | 120 | = 138x |

100 |

Selling price of Trouser = 138 x | 75 | = 103.5x |

100 |

Selling price of Shirt = 125 x | 80 | = 100x |

100 |

Difference between the selling price of Trouser and Shirt = 140

103.5x – 100x = 140

3.5x = 140

x = 40

Reqd. difference = [5 × 138 × 40 – 3 × 125 × 40]

→ 27600 – 15000 = 12600

Hence, option D is correct.

A. 12 days

B. 16 days

C. 14 days

D. 15 days

E. 20 days

A and B together can do a piece of work in 20 days.

A alone can do it in 30 days.

B alone can do the work in = | 1 | – | 1 |

20 | 30 |

B alone can do the work in 60 days.

A and C can do the work in 15 days.

C can do the work in = 30 days.

Capacity of the work = LCM {30, 60, 30} = 60 units

Units of work done A in one day = | 60 | = 2 units |

30 |

Units of work done by B in one day = | 60 | = 1 unit |

60 |

Units of work done by C in one day = | 60 | = 2 units |

30 |

Units of work done by B and C together in one day = 1 + 2 = 3 units

Units of work done by B and C together in 10 days = 3 × 10 = 30 units

Remaining work = 60 – 30 = 30 units

Time taken by A, B and C to do 30 units of work

= | 30 | = 6 days |

2 + 1 + 2 |

Total time = 10 + 6 = 16 days

Hence, option B 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

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