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- PERCENTAGE: A percentage is defined as a number represented as a fraction of 100. Percentages are used to express numbers between zero and one. It is used to compare things and use it in ratios. It is denoted by the symbol %. Percentage = (Value / Totalvalue) * 100, thus 1/ 4 = (1/4 x 100)% = 25%
- Percentage Increase/Decrease:
- If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: [R / (100 + R) x 100]%
- If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [R / (100 – R) x 100]%
Results on Population: Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
- Population after n years = P(1+ R/100)n
- Population n years ago = P / (1+ R/100)n
Results on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
- Value of the machine after n years = P (1 – R / 100)n
- Value of the machine n years ago = P /(1- R /100)n
- If A is R% more than B, then B is less than A by [R /(100 +R) x 100]%
- If A is R% less than B, then B is more than A by [R /(100 – R) x 100]%
TRUE DISCOUNT: The “True discount” is a term used when you are promised a given amount of money in the future. Since money collect interest, the actual value of that promise is less than the face value
Example:
Consider a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to R. 156 in 4 years. So, the payment of Rs. now will clear off the debt of Rs. 156 due 4 years hence. We say that:
- Sum due = Rs. 156 due 4 years hence;
- Present Worth (P.W.) = Rs. 100;
- True Discount (T.D.) = Rs. (156 – 100) = Rs. 56 = (Sum due) – (P.W.).
- We define: T.D. = Interest on P.W..; Amount = (P.W.) + (T.D.)
- Interest is reckoned on P.W. and true discount is reckoned on the amount.
IMPORTANT FORMULAS
Let rate = R% per annum and Time = T years. Then,
- P.W. = 100 x Amount / 100 + (R x T) = 100 x T.D / R x T
- T.D. = (P.W) x R x T / 100 = Amount x R x T / 100+(R x T)
- Sum = (S.I) x (T.D) / (S.I) – (T.D)
- (S.I.) – (T.D.) = S.I. on T.D.
- When the sum is put at compound interest, then P.W. = Amount / (1+R / 100)T
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Percentage and true discount - Question and Answers
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Question 1 |
Which of the following is the greatest?
40% of 30 | |
3/5 of 25 | |
6.5% of 200 | |
five more than the square of 3 |
Question 1 Explanation:
3/5 of 25b=15
40% of 30 = 12
6.5% of 200 = 13
5+9=14
40% of 30 = 12
6.5% of 200 = 13
5+9=14
Question 2 |
A new apartment complex purchased 60 toilets and 20 shower heads. If the price of a toilet is three times the price of a shower head, what percent of the total cost was the cost of all the shower heads?
9% | |
10% | |
11% | |
13% |
Question 2 Explanation:
If the cost of toilet = x & shower head = y
Total cost = 60x+20y, given x=3y
So total cost = 60*3y+20y=200y% of shower head cost to total cost = 20y/200y = 10
Total cost = 60x+20y, given x=3y
So total cost = 60*3y+20y=200y% of shower head cost to total cost = 20y/200y = 10
Question 3 |
A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
45% | |
45 5/11 % | |
54 6/11 %
| |
55% |
Question 3 Explanation:
Number of runs made by running = 110 - (3 x 4 + 8 x 6)
= 110 - (60)
= 50.
Required percentage = ( (50 x 100)/100)% = 5/4511 %
= 110 - (60)
= 50.
Required percentage = ( (50 x 100)/100)% = 5/4511 %
Question 4 |
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
39, 30 | |
41, 32 | |
42, 33 | |
43, 34 |
Question 4 Explanation:
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100 (x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.
Then, x + 9 = 56/100 (x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.
Question 5 |
A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
588 apples | |
600 apples | |
672 apples | |
700 apples |
Question 5 Explanation:
Suppose originally he had x apples.
Then, (100 - 40)% of x = 420.
60/100 x x = 420
x = (420 x 100)/60 = 700.
Then, (100 - 40)% of x = 420.
60/100 x x = 420
x = (420 x 100)/60 = 700.
Question 6 |
What percentage of numbers from 1 to 70 has 1 or 9 in the unit's digit?
1 | |
14 | |
20 | |
21 |
Question 6 Explanation:
Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69. < br/>
Number of such number =14 < br/>
Required percentage = (14/70 x 100)% = 20%
Question 7 |
If A = x% of y and B = y% of x, then which of the following is true?
A is smaller than B. | |
A is greater than B | |
Relationship between A and B cannot be determined. | |
If x is smaller than y, then A is greater than B. | |
A = B |
Question 7 Explanation:
x% of y = x/100 x y = y/100 x x = y% of x
A = B.
Question 8 |
Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
57% | |
60% | |
65% | |
90% |
Question 8 Explanation:
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 x 100) % = 57%.
Required percentage = (11628/20400 x 100) % = 57%.
Question 9 |
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
Rs. 200 | |
Rs. 250 | |
Rs. 300 | |
None of these |
Question 9 Explanation:
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
z + (120/100 )z = 550
(11/5)z = 550>br/> z = (550 x 5) 11 = 250.
Then, z + 120% of z = 550.
z + (120/100 )z = 550
(11/5)z = 550>br/> z = (550 x 5) 11 = 250.
Question 10 |
Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
Rs. 15 | |
Rs. 15.70 | |
Rs. 19.70 | |
Rs. 20 |
Question 10 Explanation:
Let the amount taxable purchases be Rs. x.
Then, 6% of x = 30
x = (30/100 * 100/6) = 5
Then, 6% of x = 30
x = (30/100 * 100/6) = 5
Question 11 |
Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
Rs. 6876.10 | |
Rs. 6999.20 | |
Rs. 6654 | |
Rs. 7000 |
Question 11 Explanation:
Rebate = 6% of Rs. 6650 = Rs.(6 x 6650) = Rs. 399.
Sales tax = 10% of Rs. (6650 - 399) = Rs. (10/100 x 6251) = Rs. 625.10
Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
Sales tax = 10% of Rs. (6650 - 399) = Rs. (10/100 x 6251) = Rs. 625.10
Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
Question 12 |
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
4.37% | |
5% | |
6% | |
8.75% |
Question 12 Explanation:
Increase in 10 years = (262500 - 175000) = 87500.
Increase% = ((87500/175000) x 100 ) % = 50%.
Required average = (50/10) % = 5%. 10
Increase% = ((87500/175000) x 100 ) % = 50%.
Required average = (50/10) % = 5%. 10
Question 13 |
If 20% of a = b, then b% of 20 is the same as:
4% of a | |
5% of a | |
20% of a | |
None of these |
Question 13 Explanation:
20% of a = b
20/100 a = b.
b% of 20 = (b/100) * 20 = (20/100)a * 1/100*20 = 4/100*a = 4% of a
20/100 a = b.
b% of 20 = (b/100) * 20 = (20/100)a * 1/100*20 = 4/100*a = 4% of a
Question 14 |
In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is of the number of students of 8 years of age which is 48. What is the total number of students in the school?
72 | |
80 | |
120 | |
150 | |
100 |
Question 14 Explanation:
Let the number of students be x. Then,
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
80% of x = 48 + 2/3 of 48
80/100 x = 80 x = 100.
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
80% of x = 48 + 2/3 of 48
80/100 x = 80 x = 100.
Question 15 |
Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
2 : 3 | |
1 : 1 | |
3:4 | |
4 : 3 |
Question 15 Explanation:
5% of A + 4% of B = 2/3 (6% of A + 8% of B)
5/100 A + 4/100 B = 2/3(6/100 A + 8/100B)
1/20A + 1/25B = 1/25A + 4/75B
(1/20-1/25)A = (4/75-1/25)B
1/100A = 1/75B
A/B = 100/75 = 4/3
Required ratio = 4:3
5/100 A + 4/100 B = 2/3(6/100 A + 8/100B)
1/20A + 1/25B = 1/25A + 4/75B
(1/20-1/25)A = (4/75-1/25)B
1/100A = 1/75B
A/B = 100/75 = 4/3
Required ratio = 4:3
Question 16 |
A student multiplied a number by 3/5 and 5/3
What is the percentage error in the calculation?
34% | |
44% | |
54% | |
64% |
Question 16 Explanation:
Let the number be x.
Then, error = 5/3 * 3/5 = 16/15
Error% =(16/15x * 3/5x * 100)% = 64%
Then, error = 5/3 * 3/5 = 16/15
Error% =(16/15x * 3/5x * 100)% = 64%
Question 17 |
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
2700 | |
2900 | |
3000 | |
3100 |
Question 17 Explanation:
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.
Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.
Question 18 |
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?
42, 33 | |
42, 36 | |
44, 33 | |
44, 36 |
Question 18 Explanation:
Let the marks secured by them be xx and (x+9)(x+9)
sum of their marks =x+(x+9)=2x+9=x+(x+9)=2x+9
Given that (x+9)(x+9) was 56% of the sum of their marks.
⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33
Then (x+9)=33+9=42(x+9)=33+9=42
Hence their marks are 33 and 42
sum of their marks =x+(x+9)=2x+9=x+(x+9)=2x+9
Given that (x+9)(x+9) was 56% of the sum of their marks.
⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33⇒(x+9)=56100(2x+9)⇒(x+9)=1425(2x+9)⇒25x+225=28x+126⇒3x=99⇒x=33
Then (x+9)=33+9=42(x+9)=33+9=42
Hence their marks are 33 and 42
Question 19 |
A true discount on an amount of Rs.480 due 3 years hence is equal as simple interest on Rs.350 for 3 years at the same interest rate on true discount. Then what is the interest rate?
11.34% | |
12.38%
| |
10.92% | |
13.12% |
Question 19 Explanation:
S.I on Rs.350 = T.D on Rs.480
This means P.W of Rs.480 due 3 years hence is Rs.350
Therefore, T.D = Rs.(480 – 350) = Rs.130
Thus S.I of Rs.350 for 3 years is Rs.130
Rate of Interest = 100 x S.I / amount x time % = 100 x 130 / 350 x 3% = 12.38%
Hence the required rate of interest is 12.38%
This means P.W of Rs.480 due 3 years hence is Rs.350
Therefore, T.D = Rs.(480 – 350) = Rs.130
Thus S.I of Rs.350 for 3 years is Rs.130
Rate of Interest = 100 x S.I / amount x time % = 100 x 130 / 350 x 3% = 12.38%
Hence the required rate of interest is 12.38%
Question 20 |
If Rs.160 and Rs.170 are true discount and simple interest of a certain amount respectively and if the time period and interest rate are equal for both simple interest and true discount then the amount is?
Rs.2120
| |
Rs.2140
| |
Rs.2720
| |
Rs.2180 |
Question 20 Explanation:
Let rate = R% per annum and Time = T years
Sum = [(S.I.) x (T.D.)]/[(S.I.) – (T.D.)] Here, S.I = 170 and T.D = 160
Then required amount = 170 x 160 / 170 – 160 = 170 x 160/10 = 2720
Hence the required sum is Rs.2720
Sum = [(S.I.) x (T.D.)]/[(S.I.) – (T.D.)] Here, S.I = 170 and T.D = 160
Then required amount = 170 x 160 / 170 – 160 = 170 x 160/10 = 2720
Hence the required sum is Rs.2720
Question 21 |
The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is?
12%
| |
13%
| |
14% | |
15% |
Question 21 Explanation:
P.W = 2562-122 = 2440
Rate = 100 * 122 / 2440 * 1/3 = 15%
Rate = 100 * 122 / 2440 * 1/3 = 15%
Question 22 |
The true discount on a certain sum due 6 months hence at 15% is Rs. 240. What is the banker's discount on the same sum for the same time at the same rate?
None of these | |
278 | |
Rs. 228 | |
Rs. 258 |
Question 22 Explanation:
TD = Rs. 240
T = 6 months = 1/2 year
R = 15%
TD = BG ×100/TR
240 = BG × 100 / [(1/2 × 15)]
BG = (240 × 15) / (100 × 2) = (120 × 15) / 100 = Rs. 18
BG = BD – TD
18 = BD - 240
BD = 18 + 240 = Rs. 258
T = 6 months = 1/2 year
R = 15%
TD = BG ×100/TR
240 = BG × 100 / [(1/2 × 15)]
BG = (240 × 15) / (100 × 2) = (120 × 15) / 100 = Rs. 18
BG = BD – TD
18 = BD - 240
BD = 18 + 240 = Rs. 258
Question 23 |
The present worth of a certain bill due sometime hence is Rs. 400 and the true discount is Rs. 20. What is the banker's discount?
Rs. 19 | |
Rs. 22 | |
Rs. 20 | |
Rs. 21 |
Question 23 Explanation:
BG = (TD)2 / PW = 202 / 400 = Rs.1
BG = BD – TD
=> 1 = BD - 20
=> BD = 1 + 20 = Rs. 21
BG = BD – TD
=> 1 = BD - 20
=> BD = 1 + 20 = Rs. 21
Question 24 |
The banker's discount on a bill due 6 months hence at 6% is Rs. 18.54. What is the true discount?
Rs. 24 | |
Rs. 12 | |
Rs. 36 | |
Rs. 18 |
Question 24 Explanation:
T = 6 months = 1/2 year
R = 6%
TD = (BD ×100)/(100+TR) = (18.54×100)/[100+(1/2 x 6)]
= (18.54×100)/103 = 1854/103 = Rs. 18
R = 6%
TD = (BD ×100)/(100+TR) = (18.54×100)/[100+(1/2 x 6)]
= (18.54×100)/103 = 1854/103 = Rs. 18
Question 25 |
The B.D. and T.D. on a certain sum is Rs.200 and Rs.100 respectively. Find out the sum?
Rs. 400 | |
Rs. 300
| |
Rs. 100 | |
Rs. 200 |
Question 25 Explanation:
F = (BD×TD)/(BD–TD) = (200×100)/(200-100) = (200x100)/100 = Rs. 200
Question 26 |
If the true discount on a certain sum due 6 months hence at 15% is Rs. 120, what is the banker's discount on the same sum for the same time and at the same rate?
50 | |
129
| |
100 | |
160 |
Question 26 Explanation:
B.G. = S.I. on T.D
=> Rs.(120 x 15 x 1/2 x 1/100)
=> Rs. 9
(B.D.) - (T.D.) = Rs. 9
B.D. = Rs. (120 + 9) = Rs. 129
=> Rs.(120 x 15 x 1/2 x 1/100)
=> Rs. 9
(B.D.) - (T.D.) = Rs. 9
B.D. = Rs. (120 + 9) = Rs. 129
Question 27 |
The true discount on a bill of Rs. 2160 is Rs. 360. What is the banker's discount?
Rs. 432 | |
Rs. 422 | |
Rs. 412 | |
Rs. 442 |
Question 27 Explanation:
F = Rs. 2160
TD = Rs. 360
PW = F - TD = 2160 - 360 = Rs. 1800
True Discount is the Simple Interest on the present value for unexpired time
=> Simple Interest on Rs. 1800 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time
= Simple Interest on Rs. 2160 for unexpired time
= (360/1800) x 2160 = (1/5) x 2160 = Rs. 432
TD = Rs. 360
PW = F - TD = 2160 - 360 = Rs. 1800
True Discount is the Simple Interest on the present value for unexpired time
=> Simple Interest on Rs. 1800 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time
= Simple Interest on Rs. 2160 for unexpired time
= (360/1800) x 2160 = (1/5) x 2160 = Rs. 432
Question 28 |
The present worth of a bill due sometime hence is Rs. 1100 and the true discount on the bill is Rs. 110. Find the banker's discount?
111 | |
121 | |
131 | |
141 |
Question 28 Explanation:
T.D. = (P.W.*B.G)(1/2)
B.G. =(T.D.)2/ P.W. = Rs.[(110x110)/ 1100] = Rs. 11
B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121
B.G. =(T.D.)2/ P.W. = Rs.[(110x110)/ 1100] = Rs. 11
B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121
Question 29 |
The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is?
Rs. 21 | |
Rs. 17 | |
Rs. 18 | |
Rs. 34 |
Question 29 Explanation:
BG = (TD)2 / PW = (340)2 /3400 = 340 x 340 / 3400 = 340/10 = Rs. 34
Question 30 |
The banker's discount on Rs. 1800 at 12% per annum is equal to the true discount on Rs. 1872 for the same time at the same rate. Find the time?
3 months | |
4 months | |
5 months | |
6 months |
Question 30 Explanation:
S.I. on Rs. 1800 = T. D. on Rs. 1872
P.W. of Rs. 1872 is Rs. 1800
Rs. 72 is S.I. on Rs. 1800 at 12%
Time = [(100x72) / (12x1800)] years = 1/3 years = 4 months
P.W. of Rs. 1872 is Rs. 1800
Rs. 72 is S.I. on Rs. 1800 at 12%
Time = [(100x72) / (12x1800)] years = 1/3 years = 4 months
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