Reveiw : Percent Increase and Decrease , Cumulative Percent , Simple and Compound Interest
Practice Examples 1, Practice Examples 2
Q.1
Quantity A Quantity B
70% of 85 85% of 70
A. Quantity A is greater
B. Quantity B is greater
C. Quantity A and Quantity B are equal
D. The relationship cannot be determined
Solution
For Quantity A; 70% of 85 = (70/100) x 85
= 59.5
For Quantity B ; 85% of 70 = (85/100) x 70
= 59.5
If we compare Quantity A with Quantity B, the values are same.
The answer is ‘C’.
Q.2
600 is what percent of 1500?
A. 40%
B. 50%
C. 20%
D. 70%
E. 55%
Solution:
Here, 100 % = 1500
1500 is 100%
So, 600 is ?
= (600 /1500) X 100
= 40%
The answer is ‘A’.
Q.3
Quantity A Quantity B
15% of 1925.25 300% of 125.25
A. Quantity A is greater
B. Quantity B is greater
C. Quantity A and Quantity B are equal
D. The relationship cannot be determined
Solution:
Let us solve for Quantity A ;
15% of 1925.25 = (15/100) x 1925.25
= 288.79
Quantity B ;
300% of 125.25 = (300/100)x 125.25
= 375.75
Quantity B is greater than Quantity A.
The answer is ‘B’.
Q.4
The students in Christ College are 23% less than Mount Carmel College. After one year 175 more students enrolled in Christ College.
Quantity A Quantity B
Total students in Christ College Total students in Mount Carmel College
A. Quantity A is greater
B. Quantity B is greater
C. Quantity A and Quantity B are equal
D. The relationship cannot be determined
Answer
Condition 1;
Suppose there are 100 students in Mount Carmel College so, 23% lesser students in Christ College i.e. 100 – 23 = 77 students in Christ College, after one year there will be 77 + 175 = 252 students in Christ College.
Here Quantity A is greater than Quantity B
Condition 2;
But if there are 800 students in Mount Carmel College then;
23% of 800 = x 800
= 184
So, in Christ College 23% lesser students compared to Mount Carmel College that means there are 800 – 184 = 616 students in Christ College. After one year 175 more students enrolled in Christ College, than students in Christ College
= 616 + 175= 791
In this condition, Quantity B is greater than Quantity A. Hence the correct answer will be ‘D’ as we cannot reach to specific conclusion with the given data.
Hence, the correct answer will be ‘D’.
Q.5
In a solution after adding chemical ‘M’ there is an increase in volume of solution by 25% and again adding chemical ‘N’ after 2 hours the volume of solution increased by 7%. What is percent increase in total volume of solution by chemical ‘M’ and chemical ‘N’?
A. 12%
B. 15%
C. 32 %
D. 33.75%
E. 35.75%
Answer;
This is based on successive percentage change.
Suppose the volume of solution is = ‘a’ = 100 ml3, we will take is as 100%, after addingchemical M , there is 25% increase in total volume of solution = 100 + 25 = 125ml3 = 125%.
Decimal equivalent of 125% = 1.25
After 2 hours again chemical ‘N’ is added so there is an increase in total volume of solution by 7%, so we have to take 7% of 125%.
We know decimal equivalent of 7% = 0.07
Now, Let us find out total percent change;
Increase in solution after adding chemical ‘N’ = 1.25 x 0.07 x a
= 0.0875 a
Percent equivalent of, 0.0875 = 8.75%.
So total percent increase in volume of solution = 125% + 8.75% = 133.75%
Change in volume of solution = 133.75 -100 = 33.75 =33.75%
OR
Increase in volume of solution = 7% of 125ml3
= 7/100 x 125
= 8.75 ml3
Total increase = 25 + 8.75 = 33.75 ml3
Actual change in volume of solution after adding chemical ‘M’ and chemical ‘N’ will be;
133.75 – 100 = 33.75 ml3
Percent Equivalent of 33.75ml³ = 33.75%|
Hence total increase in volume of solution after adding chemical ‘M’ and chemical ‘N’ will be 33.75%.
The Answer is ‘D’.