RE 5 - Arithmetic Revision Exercise | Arithmetic - Revision
All three-digit numbers, having all three digits distinct and the tens place digit as the average of the other two digits, are arranged in an ascending order to form a single number. What will be the 38th digit of that number from left?
Answer: 2
Explanation :
Initial numbers are those which starts with 1. As the middle digit is the average of other two so the last digit must be 3, 5, 7 or 9. So, the numbers are 123, 135, 147 and 159.
Similarly, the next 4 numbers, starting with 2, will be 210, 234, 246 and 258.
Next 4 numbers, starting with 3, will be 321, 345, 357 and 369.
Till now we have 12 numbers each with 3-digit, so we get first 36 digits of that number.
Now next number is 420 whose 2nd digit, which is 38th digit from the beginning, is 2.
Hence, 2.
Workspace:
On selling a certain good, the profit made by a trader is 25% of the selling price. The percentage discount he offered is same as the percentage profit earned by him. By what percentage did the trader mark up the price?
- (a)
80%
- (b)
100%
- (c)
75%
- (d)
66.67%
Answer: Option B
Explanation :
Let SP = 100
∴ Profit = 25
∴ CP = 75
∴ Profit percentage = = = 33.33% = Discount percentage
MP = 100
∴ MP = 100 × = 150
∴ MP is raised by 100%.
Workspace:
A group of girls went for shopping. The average number of items purchased by the first half of the girls is equal to the average amount spent by the other half and the average amount spent by the first half of the girls is equal to the average number of items purchased by the other half. If the total amount spent by all the girls is Rs. 234, then what can be the total number of items purchased by the girls in the first half, if the average amount spent by the girls in each half is an integer? (The average amount spent by the girls in the first half is more than Rs. 10).
- (a)
13
- (b)
6
- (c)
9
- (d)
39
Answer: Option C
Explanation :
Let the number of items purchased and the average amount spent by the first half are 'a' and 'b' respectively.
Then, the total amount spent by all the girls is 2ab. Now, if 2ab = 234 then ab = 117.
Hence a can be a factor of 117 i.e. 1, 3, 9, 13, 39 and 117. Because b is more than 10, a must be 1, 3 or 9.
Hence, option (c).
Workspace:
If the number of female employees in an office e is increased by 50% and the number of male employees is decreased by 25%, then the total number of employees in the office remains same. By what percentage will the total number of employees increase, if the number of male employees and that of female employees increases by 20% and 60% respectively?
- (a)
25%
- (b)
33.33%
- (c)
50%
- (d)
66.67%
Answer: Option B
Explanation :
Let the number of male and female employees in the office be x and y respectively.
Given that 150% of y + 75% of x = x + y
Or, 1.5y + 0.75x = x + y
Or, 0.5x = 0.25y
Or, 2y = x
Number of males after an increase of 20% = 1.2x
Number of females after an increase of 60% = 1.6y
Initial number of employees = x + y = 2y + y = 3y
Final number of employees = 1.2x + 1.6y + 1.2(2y) + 1.6y = 2.4y + 1.6y = 4y
Percentage increase = × 100 = 33.33%
Workspace:
In a class of 20 students, the average weight of the boys is 1 kg more than that of the girls while the weight of the students in the class are consecutive integers from 31 to 50 kg. If the average weight of the boys and that of the girls are integral values, then find the number of boys in the class?
Answer: 10
Explanation :
Let the number of boys be b, hence the number of girls will be 20 – b.
Let the average weight of boys be a, hence the average weight of girls will be a – 1.
⇒ b × a + (20 - b) × (a - 1) = 31 + 32 + 33 + … + 50
⇒ 20a – b = 810
⇒ a =
Since a has to be an integer, the only possible value of b = 10.
Hence, 10.
Workspace:
Siddharth, Pihu and Nidhaan paid the electricity bill of their flat for the months of January, February and March respectively. After that, Pihu and Siddharth give Rs. 250 and Rs. 450 to Nidhaan respectively, so that the final amount spent by all of them are in the ratio of 1 : 3 : 4 respectively. Find the absolute difference between the electricity bills (in Rs.) of January and February combined and March alone.
Answer: 1400
Explanation :
Let the final amount spend by three of them is 2x, 3x and 5x respectively.
Siddharth paid the bill for January and gave 450 to Nidhaan
∴ x = J + 450
⇒ J = x - 450
Pihu paid the bill for February and gave 250 to Nidhaan
∴ 3x = F + 250
⇒ F = 3x - 250
Nidhaan paid the bill for March and received a total of 700 from Siddharth and Pihu
∴ 4x = M - 700
⇒ M = 4x + 700
∴ M – (J + F) = (5x + 700) + (3x - 250) + (2x - 450)
= 1400
Hence, 1400.
Workspace:
A and B are racing on a circular track. They start simultaneously and it is known that A meets B for the first time after he finishes his 5th round and before he finishes his 6th round. If it is known that there are 4 unique points where they can meet which of the following cannot be the ratio of their speeds.
- (a)
21 : 25
- (b)
21 : 17
- (c)
23 : 19
- (d)
27 : 23
Answer: Option D
Explanation :
If there are 4 points on the circle, it must be at every quarter of the circle. So there are two possibilities:
Case 1: A is faster:
Ratio can be 5 : 4, 5 : 4 or 21 : 17, 23 : 19
Case 2: B is faster:
Ratio can be 5 : 6, 5 : 6 or 21 : 25, 23 : 27.
Hence, option (d).
Workspace:
A container contains 30 L of pure milk. From the container, 10 L milk is taken out, and replaced with 20 L of water. From the resulting mixture, 10 L mixture is taken out and replaced with 20 L of water. The same process is repeated (n – 2) more times. What is the ratio of water to milk in the final mixture?
- (a)
n(n + 5) : 6
- (b)
(n – 1) : 2
- (c)
(n – 2) : 2
- (d)
n (n + 4) : 7
Answer: Option A
Explanation :
After n such operations, the remaining quantity of pure milk in solution.
= 30 × × × … × =
The remaining quantity of water in the solution
= (30 + 10n) –
= After each operation, effectively, 10L of water is being added.
Ratio of water to milk after n such operations
= : = n(n + 5) : 6
Alternative method:
After first operation water: milk = 1 : 1.
Put n = 1 in the given options, only option (1) is satisfied.
Hence, option (a).
Workspace:
Two consecutive numbers are removed from a list of the first ‘n’ natural numbers. The average of the remaining numbers is 33 1/3. What is the product of the two numbers that have been removed?
Answer: 506
Explanation :
Average of the first ‘j’ natural numbers = .
∴ The average is either an integer or half of an integer
The final average is 33.
⇒ The number of numbers that are left must be divisible by 3.
When we erase two consecutive integers the average of the remaining numbers would either increase or decrease depending upon the numbers erased. The maximum change in the average can be +1 or -1.
So, the initial average should have been one of 32.5, 33, 33.5 or 34.
As this is the average of the first n consecutive numbers, n would be one of 64, 65, 66 or 67.
Since only 65 - 2 = 63 is divisible by 3, there were 65 numbers initially and the initial average was 33.
Sum of initial 65 numbers = 65 × 33 = 2145
Sum of remaining 63 numbers = 63 × 33 = 2100
∴ The sum of two consecutive numbers removed = 45
⇒ The two consecutive numbers are 22 and 23.
∴ Their product is 506.
Hence, 506.
Workspace:
Five men and a woman working together can do a job in 6 days. The same job is done by one woman in ‘p’ days and by eleven men in p-5 days. By what percentage is the efficiency of a woman greater than that of a man?
- (a)
300%
- (b)
500%
- (c)
600%
- (d)
700%
Answer: Option B
Explanation :
Let the efficiency of a man and a woman be ‘m’ and ‘w’ units/day respectively.
Total work done = 6 × (5m + w) = p × w = (p - 5) × 11m
⇒ =
⇒ 6 × (5p + 11(p-5)) = (p - 5) × 11p
⇒ 6 × (16p - 55) = 11p × (p - 5)
⇒ 96p - 330 = 11p2 - 55p
⇒ 11p2 – 151p + 330 = 0
⇒ p = 11 or 30/11
Since p has to be greater than 5, hence p = 11.
⇒ = = 6
⇒ Required % = × 100% = × 100% = 500%.
Hence, option (b).
Workspace:
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