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deepshikhasingh15
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Time ended
G8 Comparing quantities
Math quiz helps us to enhance brains neural network.
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1) If the ratio of boys to girls in a class is 3:5, and there are 24 boys, how many girls are there?
: Let’s use the ratio to set up an equation. Boys:Girls = 3:5. If boys = 24, then girls = (5/3) × 24 = 40.
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2) How much time will ₹7,500 take to earn ₹1,125 at 6% p.a.?
= (SI × 100) / (P × R) = (1125 × 100) / (7500 × 6) = 2.5 years
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3) A trader gives a discount of 10% on the marked price but still gains 8%. Find the marked price of an article costing ₹250.
Let MP = x SP = x × 0.90 Gain = 8% ⇒ SP = 250 × 1.08 = ₹270 So, 0.90x = 270 ⇒ x = ₹300
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4) A man buys an article at ₹960 and sells it at a gain of 20% after giving 10% discount. What is the marked price?
SP = CP × 1.20 = 960 × 1.2 = ₹1,152 Let MP be x. SP = x − 10% of x = 0.9x ⇒ 0.9x = 1152 ⇒ x = 1280 Corrected MP = ₹1280
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5) Selling price of a shirt is ₹880 with a profit of 10%. Find the cost price
CP = 880 / 1.10 = ₹800
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6) A shopkeeper earns 25% profit on selling a book at 10% discount on the marked price. If the cost price is ₹360, find the marked price.
Let MP = x, SP = 0.90x Given profit = 25% ⇒ SP = 1.25 × 360 = ₹450 So, 0.90x = 450 ⇒ x = ₹500Oops! Correct MP is ₹500 ⇒ Correct Answer: A. ₹500
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7) If 4 machines can produce 120 units in 5 hours, how many units can 6 machines produce in 5 hours?
: Let’s use the proportion: (4 machines × 120 units) = (6 machines × x units). Solving for x, we get x = 180 units.
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8) A number is first increased by 20% and then decreased by 20%. What is the net percentage change?
Net change = a + b + (ab/100) = 20 − 20 + (−400/100) = 0 − 4 = −4% ⇒ 4% decrease
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9) At what rate of interest per annum will ₹2,000 amount to ₹2,420 in 2 years under simple interest?
SI = 2420 − 2000 = ₹420 Rate = (SI × 100) / (P × T) = (420 × 100) / (2000 × 2) = 10%
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10) Find the compound interest on ₹10,000 at 8% p.a. compounded quarterly for 1 year.
Quarterly rate = 2%, n = 4 A = 10000(1 + 2/100)^4 = 10000 × (1.02)^4 ≈ ₹10824.3 CI = 824.3
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11) A shopkeeper offers 15% and 10% successive discounts on a marked price of ₹2,000. What is the final selling price?
First discount: 2000 × 0.85 = ₹1,700 Second discount: 1700 × 0.90 = ₹1,530
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12) Find the compound interest on ₹6,000 at 6.25% p.a. for 2 years compounded annually.
Amount = 6000 × (1.0625)^2 = 6000 × 1.1289 = ₹6,765.625 CI = 6765.625 − 6000 = ₹765.625
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13) Find the compound interest on ₹5,000 for 1 year at 10% per annum, compounded half-yearly.
Half-yearly rate = 5%, periods = 2 Amount = 5000 × (1 + 5/100)^2 = 5000 × 1.1025 = ₹5,512.5 CI = 5512.5 − 5000 = ₹512.5 (rounded to ₹525)
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14) A dealer marks an article 25% above cost price and allows a 20% discount. What is his gain or loss %?
Let CP = ₹100, then MP = ₹125 SP = 125 × 0.80 = ₹100 ⇒ Gain = ₹0 ⇒ Actually, it’s a 0% gain But since SP = ₹100 and CP = ₹100, no gain or loss(Note: if marked 25% above and 20% off, gain = 0)
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15) A number is increased by 25% to get 120. What is the original number?
Let’s call the original number x. Then, x + 0.25x = 120. Solving for x, we get x = 96.
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16) If the compound interest on a sum for 2 years at 12% p.a. is ₹2544, find the principal.
A = P(1.12)^2 = P × 1.2544 CI = A − P = 1.2544P − P = 0.2544P = 2544 ⇒ P = 2544 / 0.2544 = ₹10,000
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17) The ratio of ages of two persons is 4:5. After 6 years, it becomes 5:6. Find their present ages.
Solve by setting up equations.
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18) If the simple interest on ₹2,400 at 5% per annum for some years is ₹600, find the time.
SI = P × R × T / 100 ⇒ 600 = 2400 × 5 × T / 100 ⇒ 600 = 120T ⇒ T = 5 years
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19) Find the compound interest on ₹16,000 for 3 years at 5% per annum compounded annually.
A = 16000 × (1.05)^3 = 16000 × 1.157625 = ₹18,520.50 CI = 18,520.50 − 16,000 = ₹2,520.50
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20) A seller makes a profit of 15% on cost price. If the selling price is ₹460, what is the cost price?
SP = 1.15 × CP ⇒ 460 = 1.15 × CP ⇒ CP = 460 / 1.15 = ₹400
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