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## Arithmetic : Level 3 Test -8

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Question 1 |

I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins

90 | |

85 | |

100 | |

105 |

Question 1 Explanation:

The number of coins are in the ratio 2.5 : 3 : 4.,

Therefore the values will be (1 × 2.5) : (0.5 × 3) : (0.25 × 4) = 2.5 : 1.5 : 1 or 5 : 3 : 2

Therefore the total value = Rs. 210,

Let us assume the value of each coin 5r, 3r and 2r,

Therefore r = 210/10

So the total value of one-rupee coins will be 5 x (210/10) = Rs. 105

So the total number of one-rupee coins will be 105.

Therefore the values will be (1 × 2.5) : (0.5 × 3) : (0.25 × 4) = 2.5 : 1.5 : 1 or 5 : 3 : 2

Therefore the total value = Rs. 210,

Let us assume the value of each coin 5r, 3r and 2r,

Therefore r = 210/10

So the total value of one-rupee coins will be 5 x (210/10) = Rs. 105

So the total number of one-rupee coins will be 105.

Question 2 |

A man travels from A to B at a speed h km/hr. He then rests at B for h hours. He then travels from B to C at a speed 2h km/hr and rests for 2h hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be:

3 hr | |

6 hr | |

2 hr | |

4 hr |

Question 2 Explanation:

Total time taken by the man to travel from A to D = 16 hr

Total distance travelled = 36 km.

The time taken by him without taking rest

= (16 – h – 2h) = (16 – 3h).

Now the time that he take to travel individual segments

12/h + 12/2h + 12/4h = 21/h ,

Therefore,21/h = (16 - 3h)

3h

Solving this equation,

we get h = 3 or h = 7/3

This should be the time for which he rested at B.

Total distance travelled = 36 km.

The time taken by him without taking rest

= (16 – h – 2h) = (16 – 3h).

Now the time that he take to travel individual segments

12/h + 12/2h + 12/4h = 21/h ,

Therefore,21/h = (16 - 3h)

3h

^{2}– 16h + 21 = 0.Solving this equation,

we get h = 3 or h = 7/3

This should be the time for which he rested at B.

Question 3 |

A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and return at a speed 5c in the same time, then

1/a +1/b = 1/c | |

a + b = c | |

1/a +1/b = 2/c | |

none of these |

Question 3 Explanation:

Let the total distance be d.

Therefore the distance travelled by the man = 3d/5

Let the speed of man = 3a.

Therefore total time taken = 3d/15a = d/5a

Time = Distance/speed

Again distance traveled = 2d/5

Let speed = 2b.

Therefore, time taken = 2d/10b = d/5b

Total time take from A to B = d/15a + d/15b

Now he travels from B to A and comes back.

So total distance travelled = 2d

Let average speed = 5c.

Therefore time taken = 2d/ (5c),

Since the time taken in both the cases is same, we can write d/5a + d/5b = 2d/5c

Hence, 1/a + 1/b = 2/c

Therefore the distance travelled by the man = 3d/5

Let the speed of man = 3a.

Therefore total time taken = 3d/15a = d/5a

Time = Distance/speed

Again distance traveled = 2d/5

Let speed = 2b.

Therefore, time taken = 2d/10b = d/5b

Total time take from A to B = d/15a + d/15b

Now he travels from B to A and comes back.

So total distance travelled = 2d

Let average speed = 5c.

Therefore time taken = 2d/ (5c),

Since the time taken in both the cases is same, we can write d/5a + d/5b = 2d/5c

Hence, 1/a + 1/b = 2/c

Question 4 |

In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and Chinmay will win a race of one and half miles, and what will be the final lead given by the winner to the loser? (One mile is 1,600 m.)

Akshay,1/2 mile | |

Chinmay, 1/32 mile | |

Akshay, 1/24 mile | |

Chinmay, 1/16 mile |

Question 4 Explanation:

Akshay can be given a start of 128 m by Bhairav.

Therefore akshay can cover 128 m and still complete one mile with him

Akshay can travel (1600 – 128) = 1,472 m.

Therefore the ratio of the speeds of Bhairav and Akshay = Ratio of the distances travelled by them in the same time

= 1600 / 1472 = 25 : 23.

Again Bhairav can give Chinmay a start of 4 miles.

Therefore if Bhairav runs 100 m, Chinmay only runs 96 m.

So the ratio of the speeds of Bhairav and Chinmay = 100/96 = 25 : 24.

Hence, we have B : A = 25 : 23 and B : C = 25 : 24.

So A : B : C = 23 : 25 : 24.

So when Chinmay covers 24 m,

Akshay only covers 23 m.

So if they race for 1

Chinmay will complete the race first

And the distance covered by Akshay = 2,300 m.

Therefore In other words, Chinmay would beat Akshay by 100 m

= 1 /16 mile

Therefore akshay can cover 128 m and still complete one mile with him

Akshay can travel (1600 – 128) = 1,472 m.

Therefore the ratio of the speeds of Bhairav and Akshay = Ratio of the distances travelled by them in the same time

= 1600 / 1472 = 25 : 23.

Again Bhairav can give Chinmay a start of 4 miles.

Therefore if Bhairav runs 100 m, Chinmay only runs 96 m.

So the ratio of the speeds of Bhairav and Chinmay = 100/96 = 25 : 24.

Hence, we have B : A = 25 : 23 and B : C = 25 : 24.

So A : B : C = 23 : 25 : 24.

So when Chinmay covers 24 m,

Akshay only covers 23 m.

So if they race for 1

^{1}/_{2}miles = 2,400 m,Chinmay will complete the race first

And the distance covered by Akshay = 2,300 m.

Therefore In other words, Chinmay would beat Akshay by 100 m

= 1 /16 mile

Question 5 |

Ram purchased a flat at Rs. 1 lakh and Prem purchased a plot of land worth Rs. 1.1 lakh. The respective annual rates at which the prices of the flat and the plot increased were 10% and 5%. After two years they exchanged their belongings and one paid the other the difference. Then

Ram paid Rs. 275 to Prem | |

Ram paid Rs. 475 to Prem | |

Ram paid Rs. 375 to Prem | |

Prem paid Rs. 475 to Ram |

Question 5 Explanation:

The price of flat after 2 years = (1)(1.10)

Therefore the price of land =(1.1)(1.05)

Therefore, price of the plot = Rs. (1.21275 – 1.21) lakh

Which is Rs. 275 more than that of the flat.

So , Ram will have to pay Prem this amount on exchanging their belongings.

^{2}= Rs. 1.21 lakh.Therefore the price of land =(1.1)(1.05)

^{2}= Rs. 1.21275 lakh.Therefore, price of the plot = Rs. (1.21275 – 1.21) lakh

Which is Rs. 275 more than that of the flat.

So , Ram will have to pay Prem this amount on exchanging their belongings.

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