Your cart is empty.
Practice each question, then open the answer or worked solution.
A cricket bat was purchased for ₹ 800 and was sold for ₹ 1600. Then profit earned is
(a) 100% (b) 64% (c) 50% (d) 60%
Solution:-
(a) 100%
From the question it is given that,
Cost price of bat (CP) = ₹ 800
Selling price of bat (SP) = ₹ 1600
Then, Profit = SP – CP
= 1600 – 800
= ₹ 800
We know that, Profit percent = (Profit/CP) × 100
= (800/800) × 100
= 100%
A farmer bought a buffalo for ₹ 44000 and a cow for ₹ 18000. He sold the buffalo at a loss of 5% but made a profit of 10% on the cow. The net result of the transaction is
(a) loss of ₹ 200 (b) profit of ₹ 400
(c) loss of ₹ 400 (d) profit of ₹ 200
Solution:-
(c) loss of ₹ 400
From the question, it is given that,
Cost price of cow = ₹ 18000
Profit percent on cow = 10%
We know that, Profit percent = (Profit/CP) × 100
10 = (Profit/₹ 18000) × 100
(10 × 18000)/100 = Profit
Profit = 10 × 180
Profit = ₹ 1800
Then, Profit = SP – CP
1800 = SP – 18000
SP = 1800 + 18000
SP of cow = ₹ 19800
Cost price of buffalo = ₹ 44000
He sold the buffalo at a loss of 5%
We know that, Loss percent = (loss/CP) × 100
5 = (loss/44000) × 100
(5 × 44000)/100 = loss
Loss = 5 × 440
Loss = ₹ 2200
Loss = CP – SP
₹2200 = ₹44000 – SP
SP = 44000 – 2200
SP of buffalo = ₹ 41,800
Now,
Total cost price of both buffalo and cow = ₹ 44000 + ₹ 18000
= ₹ 62000
Total selling price of both buffalo and cow = ₹ 41800 + ₹ 19800
= ₹ 61,600
Selling price of both buffalo and cow is less when compared to the cost price,
So, Loss = CP – SP
= ₹ 62000 – ₹ 61600
Loss = ₹ 400
If Mohan’s income is 25% more than Raman’s income, then Raman’s income is less than Mohan’s income by
(a) 25% (b) 80% (c) 20% (d) 75%
Solution:-
(c) 20%
Let us assume Mohan’s income be ‘P’ and Raman’s income be ‘Q’
From the question it is given that,
Mohan’s income is 25% more than Raman’s income i.e. x = y + 25% of y
X = y + (25/100) y
X = y (1 + (25/100))
X = y ((100 + 25)/100)
X = y(125/100)
Then, the difference between both the income is = (125/100)y – (25/100)y
So,
Raman’s income as compared to Mohan’s income,
= (difference)/ (Mohan’s income) × 100
= ((25y/100)/(125y/100)) × 100
= (25/125) × 100
= 20%
The interest on ₹ 30000 for 3 years at the rate of 15% per annum is
(a) ₹ 4500 (b) ₹ 9000 (c) ₹ 18000 (d) ₹ 13500
Solution:-
(d) ₹ 13500
From the question, it is given that,
Principal= ₹ 30000
Time = 3 Years
Rate = 15%
Then, we know the formula of Simple interest I = (P × R × T)/100
I= (30000 × (15) × 3)/100
I = (30000 × 15 × 3)/100
I = ₹ 13500
Amount received on ₹ 3000 for 2 years at the rate of 11% per annum is
(a) ₹ 2340 (b) ₹ 3660 (c) ₹ 4320 (d) ₹ 3330
Solution:-
(b) ₹ 3660
From the question, it is given that,
Principal= ₹ 3000
Time = 2 Years
Rate = 11%
Then, we know the formula of Simple interest I = (P × R × T)/100
I= (3000 × (11) × 2)/100
I = (3000 × 11 × 2)/100
I = ₹ 660
Amount = P + I
= 3000 + 660
= ₹ 3660
Interest on ₹ 12000 for 1 month at the rate of 10 % per annum is
(a) ₹ 1200 (b) ₹ 600 (c) ₹ 100 (d) ₹ 12100
Solution:-
(c) ₹ 100
From the question it is given that,
Principal= ₹ 12000
Time = 1 month = 1/12
Rate = 10%
Then, we know the formula of Simple interest I = (P × R × T)/100
I= (12000 × (10) × (1/12))/100
I = ₹ 100
Rajini and Mohini deposited ₹ 3000 and ₹ 4000 in a company at the rate of 10% per annum for 3 years and 2½ years respectively. The difference of the amounts received by them will be
(a) ₹ 100 (b) ₹ 1000
(c) ₹ 900 (d) ₹ 1100
Solution:-
(d) ₹ 1100
From the question it is given that,
Money deposited by Rajini = ₹ 3000
Time = 3 years
Rate = 10%
Then, we know the formula of Simple interest I = (P × R × T)/100
I= (3000× (10) × 3)/100
I = ₹ 900
Amount = P + I
= 3000 + 900
= ₹ 3900
Now,
Money deposited by Mohini = ₹ 4000
Time = 2½ = 5/2
Rate = 10%
Then, we know the formula of Simple interest I = (P × R × T)/100
I= (4000× (10) × (5/2))/100
I = ₹ 1000
Amount = P + I
= 4000 + 1000
= ₹ 5000
The difference of the amounts received by them will be = 5000 – 3900
= ₹1100
If 90% of x is 315 km, then the value of x is
(a) 325 km (b) 350 km (c) 405 km (d) 340 km
Solution:-
(b) 350 km
From the question it is given that,
90% of x is 315 km, it can be written as, (90/100) × x = 315
x = (315 × 100)/90
x = 31500/90
x = 350 km
On selling an article for ₹ 329, a dealer lost 6%. The cost price of the article is
(a) ₹ 310.37 (b) ₹ 348.74 (c) ₹ 335 (d) ₹ 350
Solution:-
(d) ₹ 350
From the question it is given that,
Selling price of the article = ₹ 329
Loss percent = 6%
We know that, Loss percent = (loss/CP) × 100
6 = ((CP – SP)/CP) × 100
(6/100)CP = CP –SP
CP – (6/100)CP = 329
(94/100)CP = 329
CP = (329 × 100)/94
CP = ₹ 350
(25% of 50% of 100%)/(25 × 50) is equal to
(a) 1.1% (b) 0.1% (c) 0.01% (d) 1%
Solution:-
(c) 0.01%
= [(25/100) × (50/100) × (100/100)]/ (25 × 50)
= [¼ × ½ × 1]/(1250)
= [1/8] × (1/1250)
= 1/10000
= 0.0001 × 100
= 0.01%