| 1. x varies directly as the cube of y. When y = 11, x is 3993. Find x when y = 21. | Easy |
| A. 27783 B. 18522 C. 37044 D. None of these |
View Answer
Answer: Option A
Explanation:
x ∞ y3 or x = ky3
-> 3993 = k (11)3 or k = 3993/1331 = 3
x = 3y3
When y = 21, x = 3*213 = 27783
| 2. The relationship between the volume of a solid and its dimensions can be described as follows: when the height remains constant, the volume is directly proportional to the square of the radius; and when the radius remains constant, the volume is directly proportional to the height. Given that the radius of a solid is 5 units and the height is 21 units, the volume is measured to be 550. Find the volume of the solid when the radius is 7 cm, and the height is 42 cm. | Easy |
| A. 4312 cm3 B. 1078 cm3 C. 2156 cm3 D. None of these |
View Answer
Answer: Option C
Explanation:
V ∞ r2, V ∞ h
-> V ∞ r2h
-> V = kr2h
Putting values, we get 550 = k (5)2 *21
-> k = 550/525 = 22/21
-> V = (22/21)*72*42 = 2156
| 3. The value of a product is directly proportional to the square root of its diameter when the thickness is constant; the value is directly proportional to the square of thickness when the diameter is constant. The value of one such product is four times that of a second, and their diameters are in the ratio 4: 1, respectively. Find the ratio of their thickness. | Medium |
| A. 1: 21/2 B. 2:1 C. 21/2: 1 D. 1:4 |
View Answer
Answer: Option C
Explanation:
Given V ∞ d1/2 and V ∞ t2
-> V ∞ t2d1/2
-> V1 = kt12d11/2 and V2 = kt22d21/2
-> V1/V2 = kt12d11/2 / kt22d21/2 = 4
-> (t1/t2)2(4d/d)1/2 = 4
-> t1/t2 = 21/2:1
| 4. The ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant. Find the approximate volume of a gas at STP when 2 litres are collected at 745 mm Hg and 298 Kelvin. STP refers to “standard temperature and pressure,” which is 273 Kelvin and 760 mm Hg, respectively. | Easy |
| A. 2.8 L B. 1.8 L C. 0.8 L D. None of these |
View Answer
Answer: Option B
Explanation:
Given that the product of pressure and volume and the absolute temperature of a gas is equal to a constant -> PV/T = k
-> P1V1 / T1 = P2V2 / T2
->P1V1T2 = P2V2T1
Putting values, P1 = 745 mm Hg, V1 = 2 L, T1 = 298 K, P2 = 760 mm Hg, V2 = x (the unknown you’re solving for), T2 = 273 K
-> (745*2*273) = (760*x*298)
-> x = 1.8 L
| 5. The relationship between the length of a pendulum and its frequency is such that the length is inversely proportional to the square of the frequency. Given that a pendulum with a length of 12 ft. oscillates 24 times per minute determine the length of a pendulum that oscillates 16 times per minute. | Easy |
| A. 27 ft. B. 20 ft. C. 30 ft. D. None of these |
View Answer
Answer: Option A
Explanation:
L = k*1/f2
Now 12 = k*1/242
-> k = 12*242
Therefore, L2 = 12*242 *1/162 = 27 ft.
| 6. The price of product A varies as the square root of its weight. One such product weighing 16 units costs Rs. 200. The price of product B varies as the square of its weight. One such product weighing 16 units costs Rs. 6400. Both A and B fell and broke into two pieces, each weighing 4 units and 12 units. Which product experiences a greater percentage change in value after breakage compared to the unbroken product, and how much? | Difficult |
| A. Product A, 36.6% B. Product B, 37.5% C. Product B, 60% D. None of these |
View Answer
Answer: Option B
Explanation:
For A, Price ∞ (Weight)1/2
-> PA = kA*wA1/2, where kA is a constant for product A.
-> 200 = kA* 161/2
-> kA = 50
After breakage, new price = (50*41/2) + (50*121/2) = 50 (2+2.31/2) = 273.2
-> Absolute change in value = Rs. 73.2
% change = (73.2/200) * 100 = 36.6%
For B, Price ∞ (Weight)2
-> PB = kB*wB2, where kB is a constant for product B.
-> 6400 = kB * 162
-> kB = 25
After breaking into pieces, new price = (25*42)+(25*122) = 4000
-> Absolute change in value = 6400-4000 = Rs. 2400
% change = (2400/6400) *100 = 37.5%
Thus, product B experiences a higher % change in value.
| 7. Consider a scenario where the quantity “b” changes based on the sum of two components. One of these components varies directly with “x,” while the other component varies inversely with “x.” Given that when “x” is equal to 4, “b” is 6, and when “x” is equal to 3, “b” is 10/3, what would be the value of “b” when “x” is equal to 8? | Medium |
| A. 15 B. 12 C. 10 D. None of these |
View Answer
Answer: Option A
Explanation:
b = k1x + (k2/x), where k1, k2 are constants.
Putting b = 6 and x = 4,
We get, 16k1 + k2 = 24 ……(i)
Putting b = 10/3 and x = 3,
We get, 9k1 + k2 = 10 ……(ii)
Solving (i) and (ii), we get, k1 = 2, k2 = -8
∴ b = 2x – 8/x
When x = 4, b = 2(8) -8/8 = 15
| 8. The volume of a solid is directly proportional to the square of its radius (when the height is constant) and directly proportional to the height (when the radius is constant). What is the percent change in volume if the radius is increased by 50% and the height is reduced by 50%? | Easy |
| A. 12.5% increase B. 12.5% decrease C. 10% increase D. Cannot be determined |
View Answer
Answer: Option A
Explanation:
V ∞ r2, V ∞ h
-> V ∞ r2h
-> V = kr2h
New volume = VN = k (1.5r)2(0.5h) = 1.125V
% change = ((1.125V-V)/V) * 100 = 12.5%
| 9. The volume (V) of a given mass of gas varies directly with the gas’s absolute temperature (T) when pressure is kept constant. An open “empty” 5 L plastic pop container with an actual inside volume of 5.05 L is removed from a refrigerator at 280 K and allowed to warm up to 300 K. What air volume, measured at 300 K, will leave the container as it warms? | Medium |
| A. 0.41 L B. 0.36 L C. 0.26 L D. None of these |
View Answer
Answer: Option B
Explanation:
V ∞ T or V = kT
-> V1/T1 = V2/T2
-> 5.05/280 = V2/300
-> V2 = 5.41 L
The volume that “escapes” is V2 minus 5.05 L.
Required volume = 5.41 – 5.05 = 0.36 L
| 10. The price of a precious stone is directly proportional to the square of its weight. Unfortunately, one particular stone accidentally broke into four pieces, with weights in the ratio of 2:3:5:10. The total loss in price due to this accident amounts to Rs. 91700. Determine the price of the original piece of the stone. | Difficult |
| A. Rs. 140000 B. Rs. 120000 C. Rs. 100000 D. None of these |
View Answer
Answer: Option A
Explanation:
P = k*w2
Where P is the price and w is the weight of the stone.
The weights of the pieces are in the ratio 2: 3: 5: 10
-> The weights of the broken pieces are 2w/20,3w/20,5w/20 & 10w/20 respectively.
-> P1 = k(2w/20)2, P2 = k(3w/20)2, P3 = k(5w/20)2, P4 = k(10w/20)2
Therefore, total price = P1+P2+P3+P4 = kw2(138/400) = kw2(69/200)
Original Price = kw2, Price after breakage = kw2
Loss = kw2 – kw2 = kw2 = 91700
-> kw2 = 140000 (original price)