- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Arithmetic : Level 3 Test -8

Congratulations - you have completed

*Arithmetic : Level 3 Test -8*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

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.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |