Olympiad Mathematics - Comparing Quantities
Exam Duration: 45 Mins Total Questions : 25
When the price of a product was decreased by 10%, the number of products sold increased by 30%. What was the increase on the total revenue?
- (a)
23%
- (b)
16\(\frac{2}{3}\)%
- (c)
15%
- (d)
17%
Let the price of the product be Rs 100 and let original sale of product = 100. Then,
Total revenue = Rs (100 x 100) = Rs 10000
Now, the price is decreased by 10%
∴. New price = Rs 90
Also, number of pieces sold is increased by 30%.
∴ Total pieces sold = 130
∴ New revenue = Rs(130 x 90) = Rs 11700
Now, increase in revenue
=\(\left( \frac { (11700-10000) }{ 10000 } \times 100 \right) \)=17%
If a number x is 10% less than the another number y and y is 10% more than 125, then x is equal to _____
- (a)
123.75
- (b)
140.55
- (c)
143
- (d)
150
It is given that, x=y-10% of y ⇒ x=\(\frac{9}{10}\)y
Also, y=125+10% of 125 =125+12.5 ⇒ y=137.5
∴ x=\(\frac{9}{10}\) x 137.5 ⇒ x=123.75
If 23% of 8 is 46, then find 8.
- (a)
150
- (b)
200
- (c)
20
- (d)
300
We have, 23% of a = 46
⇒ \(\frac{23}{100}\) x a =46 ⇒ a=46 x \(\frac{100}{23}\)=200
A number is increased by 10% and then it is decreased by 10%. Find the net increase or decrease percent.
- (a)
Decrease by 1%
- (b)
Decrease by 10%
- (c)
Increase by 2%
- (d)
Increase by 11%
Let the number be 100
Increased number = 100 + 10% of 100 = 110
Now, this number is decreased by 10%
∴ Newnumber=110-10% of 110 = 99
Net decrease = 100 - 99 = 1
∴ Net percentage decrease =\(\left( \frac { 1 }{ 100 } \times 100 \right) \)% =1%
If S.P. of an article is \(\frac{4}{3}\) of its C.P., then the profit % in the transaction is ______
- (a)
\(\frac{1}{3}\)%
- (b)
20\(\frac{1}{2}\)%
- (c)
33\(\frac{1}{3}\)%
- (d)
25\(\frac{1}{2}\)%
Let C.P. of an article be Rs x
∴ S.P. of the article = Rs \(\left( \frac { 4 }{ 3 } x \right) \)
Profit = S.P -C.P = Rs \(\left( \frac { 4 }{ 3 } x-x \right) \)=Rs \(\left( \frac { 1 }{ 3 } x \right) \)
∴ Profit % = \(\frac { Profit }{ C.P } \times 100=\frac { \frac { 1 }{ 3 } x }{ x } \times 100\)=33\(\frac{1}{3}\)%
If 30% of 140 = x% of 840, then the value of x is ______
- (a)
5
- (b)
15
- (c)
24
- (d)
60
We have, 30% of 140 = x% of 840
⇒ \(\frac { 30 }{ 100 } \times 140=\frac { x }{ 100 } \times 840\Rightarrow 42=\frac { 42 }{ 5 } \)x ⇒ x=5
x is 5% of y, y is 24% of z. If x = 480, find the values of y and z respectively.
- (a)
9500, 40000
- (b)
9600, 40000
- (c)
9800, 50000
- (d)
9600, 50000
We have, x = 480
According to question, x is 5% of y
⇒ 480 =\(\frac { 5 }{ 100 } \) x y ⇒ y=9600
Also, y is 24% of z
⇒ 9600 =\(\frac{24}{100}\) x z ⇒ z=40000
If 12.5% of 192 = 50% of x, then x =
- (a)
45
- (b)
25
- (c)
48
- (d)
92
We have, 12.5% of 192 = 50% of x
⇒ \(\frac { 12.5 }{ 100 } \times 192=\frac { 50 }{ 100 } \times x\) ⇒ x=24 x 2 ⇒ x=48.
The simple interest at x % for x years will be Rs x on a sum of ______
- (a)
Rs x
- (b)
Rs 100x
- (c)
Rs \(\left( \frac { 100 }{ x } \right) \).
- (d)
Rs \(\left( \frac { 100 }{ x^{ 2 } } \right) \).
S.I. = Rs x, Rate = x%, Time = x years,
S.I = \(\frac { P\times R\times T }{ 100 } \Rightarrow x=\frac { P\times x\times x }{ 100 } \) ⇒ P=Rs \(\left( \frac { 100 }{ x } \right) \).
Simple interest on a certain amount is \(\frac{9}{16}\) of the principal. If the numbers representing the rate of interest (in percent) and time (in years) be equal, then time for which the principal is lent out, is ________
- (a)
5\(\frac{1}{2}\) years
- (b)
6\(\frac{1}{2}\) years
- (c)
7 years
- (d)
7\(\frac{1}{2}\) years
Let the principal be Rs P
And the rate of interest: Time: n (say)
According to question,
S.I =\(\frac{9}{16}\)P ⇒ \(\frac { P\times n\times n }{ 100 } =\frac { 9 }{ 16 } \)P
⇒ n2=\(\frac { 9 }{ 16 } P\times \frac { 100 }{ P } \Rightarrow n=\frac { 15 }{ 2 } \Rightarrow n=7\frac { 1 }{ 2 } \) years.
A watch worth Rs 5400 is offered for sale at Rs 4500. What percent discount is offered during the sale?
- (a)
50/3
- (b)
49/3
- (c)
25/3
- (d)
34/3
We have, M.P. = Rs 5400 and S.P. = Rs 4500
∴ Discount: M.P. - S.p. =Rs 5400 -Rs 4500 = Rs 900
Thus, Discount % = \(\left( \frac { Discount }{ M.P } \times 100 \right) =\left( \frac { 900 }{ 5400 } \times 100 \right) =\frac { 50 }{ 3 } \).
The population of a town was decreasing every year due to migration, poverty and unemployment. The present population of the town is 6,31,680. Last year the migration was 4% and the year before last, it was 6%. What was the population two years ago?
- (a)
9,00,000
- (b)
5,00,000
- (c)
6,00,000
- (d)
7,00,000
Let the population two years ago= x
After 1 year, it remained = x-\(\frac { 6 }{ 100 } \)x
After 2 years, it remained = \(\left( x-\frac { 6x }{ 100 } \right) -\left( x-\frac { 6x }{ 100 } \right) \times \frac { 4 }{ 100 } \)
Since, present population: population after 2 years
⇒ 631680 = \(x-\frac { 6x }{ 100 } -\frac { 4x }{ 100 } +\frac { 24x }{ 10000 } \)
⇒ 631680 =\(\frac { 10000x-600x-400x+24x }{ 10000 } \)
⇒ 631680 =\(\frac { 9024x }{ 1000 } \Rightarrow \frac { 631680\times 10000 }{ 9024 } \)=x
⇒ x=700000
Monika purchased a pressure cooker at \(\left( \frac { 9 }{ 10 } \right) ^{ th }\) of its selling price and sold it at 8% more than than its S.P. Find her gain percent.
- (a)
20%
- (b)
10%
- (c)
30%
- (d)
40%
Let the selling price be Rs 100
Then, cost price of cooker = \(\left( \frac { 9 }{ 10 } \times 100 \right) \) = Rs 90
and selling price of cooker for Monika
= Rs 100 + 8% of Rs 100 =Rs 108
∴ Gain % =\(\left( \frac { 108-90 }{ 90 } \right) \) x 100 = 20%
Sam invested Rs 15000 at 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be
- (a)
Rs 16500
- (b)
Rs 16525.50
- (c)
Rs 16537.50
- (d)
Rs 18150
Principal: Rs 15000
Rate: 10% p.a. = 5% half yearly
Time = 1 year = 2 half years
∴ Amount = 15000\(\left( 1+\frac { 5 }{ 100 } \right) ^{ 2 }\) = Rs 16537.50.
A real estate agent receives Rs 50,000 as commission, which is 4% of the selling price. At what price does the agent sell the property?
- (a)
Rs 1250000
- (b)
Rs 1025000
- (c)
Rs 1125000
- (d)
Rs 1450000
Let the selling price of the property: Rs x
Rate of commission: 4% on selling price = Rs \(\left( \frac { 4 }{ 100 } x \right) \)
Since, a real estate agent receives Rs 50000
∴ Rs \(\left( \frac { 4 }{ 100 } x \right) \) = Rs⇒ x= 50000 x \(\frac { 100 }{ 4 } \) =1250000
∴ The selling price of the property is Rs 1250000.
Abha purchased a house from Avas Parishad credit. If the cost of the house is Rs 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one and a half year.
- (a)
4900
- (b)
Rs 4921
- (c)
Rs 4810
- (d)
Rs 4700
We have, Principal = Rs 64000
Rate: 5% per annum = \(\frac { 5 }{ 2 } \)% per half yearly
Time =1\(\frac { 1 }{ 2 } \) years = \(\left( \frac { 3 }{ 2 } \times 2 \right) \) half years: 3 half years.
Now, Amount = 64000 \(\left( 1+\frac { 5 }{ 200 } \right) ^3\) ⇒ Amount = Rs 68921
Compound interest: 68921 - 64000 = Rs 4921.
Select the incorrect match.
- (a)
Principal(In Rs) Rate %(p.a) Time (in years) C.I. (In Rs) 1000 5% 3 157.63 - (b)
Principal(In Rs) Rate %(p.a) Time (in years) C.I. (In Rs) 360 20% 2 158.40 - (c)
Principal(In Rs) Rate %(p.a) Time (in years) C.I. (In Rs) 3000 10% 3 939 - (d)
Principal(In Rs) Rate %(p.a) Time (in years) C.I. (In Rs) 72000 6% 3 13753.15
Principal = Rs 3000
Rate: 10%, Time = 3 years
A=P \(\left( 1+\frac { R }{ 100 } \right) ^{ n }=3000\left( 1+\frac { 10 }{ 100 } \right) ^{ 3 }\)=Rs 3993
C.I = Rs (3993 - 3000) = Rs 993.
Pankaj borrowed Rs 8000 for 2 year at 15% per annum. Calculate the amount, if interest is
(i) compounded annually
(ii) compounded half yearly
- (a)
(i) (ii) Rs 10550 Rs 11883.2 - (b)
(i) (ii) Rs 10090 Rs 12432.75 - (c)
(i) (ii) Rs 10580 Rs 10502.50 - (d)
(i) (ii) Rs 10580 Rs 10683.75
The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 2 years us Rs 500. Find the sum when the interest compounded annually.
- (a)
Rs 50000
- (b)
Rs 55000
- (c)
Rs 40000
- (d)
Rs 65000
Let P be the principal.
According to question, C.I.- S.I. =Rs 500
⇒ \(\left[ P\left( 1+\frac { 10 }{ 100 } \right) ^{ 2 }-P \right] -\left[ \frac { P\times 10\times 2 }{ 100 } \right] \)
⇒ \(\frac { 21 }{ 100 } P-\frac { 1 }{ 5 } \)P =500
⇒ \(\frac { P }{ 100 } \) =500 ⇒ P=500 x 100 = Rs 50000
State 'T' for true and 'F' for false.
(i) A shopkeeper bought a cycle for Rs 1200 and sold it for Rs 1500, then his gain percentage is 25%
(ii) 200 kg of sugar was purchased at the rate of Rs 15 per kg and sold at a profit of 5%. Then selling price of sugar is Rs per kg.
(iii) A person sells an article for Rs 550 and gain \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the cost price. Then the gain percent is 11%.
(iv) The cost price of a dinning table is Rs 1500 and its marked price is Rs 1800. If a shopkeeper sells it at a loss of 8% then the discount offered by him is 23\(\frac{1}{3}\) .
- (a)
(i) (ii) (iii) (iv) T T F T - (b)
(i) (ii) (iii) (iv) F F T F - (c)
(i) (ii) (iii) (iv) T F F T - (d)
(i) (ii) (iii) (iv) F F T T
(i) C.P. = Rs 1200, S.P. = Rs 1500
∴ Gain % =\(\frac { (1500-1200) }{ 1200 } \times 100=\frac { 300 }{ 1200 } \times 100\) =25%
(ii) C.P. of 1 kg of sugar = Rs 15
C.P. of 200 kg of sugar = Rs (200 x 15) = Rs 3000
Profit = 5%
∴ S.P. = \(\frac{105}{100}\times\)3000 =3150
∴ S.P of 1 kg of sugar = \(\frac{3150}{200}\)=Rs 15.75
(iii) S.P. of article = Rs 550
Gain = \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of C.P. ⇒ Gain % = \(\frac { \left( \frac { 1 }{ 10 } \right) \times C.P }{ C.P } \times 100\)=10%
(iv) C.P. of dinning table = Rs 1500 Marked price = Rs 1800 Loss = 8%
S.P. = \(\frac { 1500\times 92 }{ 100 } \) = Rs 1380
Now, Discount = 1800-1380= Rs 420
∴ Discount % = \(\frac { 420 }{ 1800 } \) x 100 = 23\(\frac{1}{3}\)%.
Match the following:
Column-I | Column-II |
Gun power contains 75% nitre and 10% sulphur. The rest of it is charcoal. The amount of charcoal in 9 kg of gun powder (in kg) is |
1.5 |
(ii) A cycle merchant allows 25% discount on the marked price of the cycle and still makes a profit of 20% If he gains Rs 360 over the sale of one cycle, the marked price of the cycle (in Rs ) is |
1.35 |
(iii) Time (in years) in which Rs 64000 will amount of Rs 68921 at 5% p.a. interest being compounded semi-annually is |
2880 |
- (a)
(i)⟶(c), (ii)⟶(b), (iii)⟶(a)
- (b)
(i)⟶(b), (ii)⟶(a), (iii)⟶(c)
- (c)
(i)⟶(b), (ii)⟶(c), (iii)⟶(a)
- (d)
(i)⟶(c), (ii)⟶(a), (iii)⟶(b)
(i) Total quantity of gun powder = 9 kg
Quantity of nitre = 75% of 9 kg = 6.75 kg
Quantity of sulphur = 10% of 9 kg = 0.9 kg
∴ Quantity of charcoal = [9 - (6.75 + 0.90)]kg = 1.35 kg
(ii) Let the marked price be Rs x
Then, S.P = 75% of x=\(\left( \frac { 3 }{ 4 } x \right) \)
According to question, C.P. =\(\frac { \frac { 3 }{ 4 } x\times 100 }{ (20+100) } =\frac { 5 }{ 8 } x\)
Now ,gain = RS 360 ⇒ \(\frac { 3 }{ 4 } \)x-\(\frac { 5 }{ 8 } \)x=360 ⇒ x=2880
So, marked price = Rs 2880
(iii) P = Rs 64000, A = Rs 68921
R=5% p.a. =\(\frac{5}{2}\)% per half year
Let time = n years = 2n half years
∴ 68921 = 64000\(\frac { 68921 }{ 64000 } =\left( \frac { 41 }{ 40 } \right) ^{ 2n }\) ⇒ \(\left( \frac { 41 }{ 40 } \right) ^{ 3 }=\left( \frac { 41 }{ 40 } \right) ^{ 2n }\)
On comparing, 2n = 3 ⇒ n = 1.5 years.
A grocer purchased 80 kg of sugar at Rs 13.50 kg and mixed it with 120 kg of sugar of cost Rs 16 per kg. At what rate should he sell the mixture (per kg) to gain 16%?
- (a)
Rs 15.30
- (b)
Rs 19.18
- (c)
Rs 17.40
- (d)
Rs 18.66
Cost price of 80 kg of Isltype sugar
=Rs (13.50 x 80) Rs 1080
Cost price of 120 kg of Iindtype sugar
=Rs (16 x 120) =Rs 1920
∴ Total cost price : Rs (1920 + 1080) =3000
Gain% = 16%
Now, S.P. = \(\left( \frac { 100+Gain% }{ 100 } \right) \times C.P=\frac { 116 }{ 100 } \times 3000\) =Rs 3480
i.e. Total selling price of 200 kg sugar = Rs 3480
∴ S.P. of 1kg sugar = Rs \(\left( \frac { 3480 }{ 200 } \right) \) = Rs 17.40
A milkman sold two of his buffaloes for Rs 20000 each. On one he made a gain of 5% and on the other a loss of 10% Find his overall gain or loss.
- (a)
Loss of Rs 1269.84
- (b)
Gain of Rs 4268.84
- (c)
Gain of Rs 1269.84
- (d)
Loss Rs 1200
S.P. of 1st buffalo: Rs 20000
Gain%: 5%
∴ C.P. =\(\frac { 100\times 20000 }{ 105 } \) = Rs 19047.62
S.P. of 2nd buffalo = Rs 20000 and loss % =10%
∴ C.P. = \(\frac { 100\times 20000 }{ 90 } \) = Rs 22222.22
Total C.P. = Rs (19047.62 + 22222.22): =Rs 41269.84
Total S.P. = Rs 40000
∴ Loss: C.P. - S.P.: =Rs (41269.84 - 40000) =Rs 1269.84
A book was sold for Rs 27.50 with a profit of 10%. If it were sold for Rs 25.75 then what would have been the percentage of profit or loss?
- (a)
Loss, 3%
- (b)
Profit, 2%
- (c)
Profit, 3%
- (d)
Loss, 2%
Selling price of book = Rs 27.50
Profit % = 10%
∴ C.P. = \(\frac{100}{110}\) x 27.50 = Rs 25
Now, new selling price = Rs 25.75
∴ Profit = Rs (25.75 - 25) = Rs0.75
Now, Profit % = \(\frac{0.75}{25}\) x 100 =3%.
The cost of a vechicle is Rs 175000. If its value depreciates at the rate of 20% per annum then the total depreciation after 3 years was _______
- (a)
Rs 86400
- (b)
Rs 82500
- (c)
Rs 84500
- (d)
Rs 85400
Cost of vehicle =Rs175000
Rate of depreciation =20% p.a.
∴ Depreciated value after 3 years
=175000\(\left( 1-\frac { 20 }{ 100 } \right) ^{ 3 }=175000\times \left( \frac { 8 }{ 10 } \right) ^{ 3 }\)= Rs 89600
∴ Total depreciation = Rs (175000 - 89600) = 85400.