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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)
3h2 – 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.
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)
3h2 – 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 11/2miles = 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
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 11/2miles = 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)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.
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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