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Browse complete papers45 - [38 - {80 ÷ 4 - (8 - 12 ÷ 3) ÷ 4}] = ?
25
27
26
28
Correct answer
26
Why this answer is correct
Calculation:
45 - [38 - {80 ÷ 4 - (8 - 12 ÷ 3) ÷ 4}]
⇒ 45 - [38 - {20 - (8 - 4) ÷ 4}]
⇒ 45 - [38 - 19]
⇒ 45 - 19
⇒ 26
The Table shown the Number of Laptops and Desktops manufactured by a company.
| Year | 2008 | 2009 | 2010 | 2011 | 2012 |
|---|---|---|---|---|---|
| Laptops | 14400 | 20500 | 12800 | 16400 | 18600 |
| Desktops | 12800 | 24700 | 19200 | 20200 | 14900 |
What is the average number of laptops manufactured by a company from 2008 to 2012?
13550
16540
14260
16140
Correct answer
16540
Why this answer is correct
Total number of laptops manufactured by a company from 2008 to 2012 = 14400 + 20500 + 12800 + 16400 + 18600
⇒ 82700
∴ The average number of laptops manufactured by a company from 2008 to 2012 = 82700/5
⇒ 16540
Pipes M and N can alone fill and alone empty the tank in 12 hours and 24 hours respectively. There is a leakage at the bottom of tank which can empty the tank in 72 hours. If all pipes opened together, then in how much time tank will filled?
48 hours
30 hours
36 hours
24 hours
Correct answer
36 hours
Why this answer is correct
Shortcut Trick
Assume Total Capacity of tank = LCM of (12, 24, 72) = 72 units.
Efficiency of Pipe M = 72 ÷ 12 = +6 | Pipe N = 72 ÷ 24 = −3 | Leakage = 72 ÷ 72 = −1.
Combined Efficiency of all together = 6 − 3 − 1 = 2 units/hr.
Time required to fill the tank = 72 ÷ 2 = 36 hr.
∴ The correct answer is 36 hr.
Alternate Method
Given:
Time taken by Pipe M to fill = 12 hr
Time taken by Pipe N to empty = 24 hr
Time taken by Leakage to empty = 72 hr
Formula Used:
1/Ttotal = 1/TM − 1/TN − 1/TLeak
| Person/Pipe | Time (hr) | Total Work (units) | Efficiency (units/hr) |
|---|---|---|---|
| Pipe M (Inlet) | 12 | 72 | +6 |
| Pipe N (Outlet) | 24 | −3 | |
| Leakage (Outlet) | 72 | −1 |
Calculations:
⇒ Total efficiency when all are open = 6 + (−3) + (−1)
⇒ Total efficiency = 6 − 4 = 2 units/hr
⇒ Time taken to fill the tank = Total Capacity ÷ Total Efficiency
⇒ Time = 72 ÷ 2
⇒ Time = 36 hr
∴ The correct answer is 36 hr.
Additional Information
Inlet and Outlet Pipes
An inlet pipe fills the tank (positive work), while an outlet pipe or leakage empties the tank (negative work).
Efficiency Rule
If a pipe fills a tank in 'x' hours, its 1-hour work is 1/x. If another empties it in 'y' hours, net work in 1 hour is (1/x − 1/y).
Combined Time Formula
For two filling pipes A and B, total time = (A × B) ÷ (A + B). For one filling (A) and one emptying (B), total time = (A × B) ÷ (B − A).
‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
\(10\frac{1}{2} \) minutes
25 minutes
15 minutes
4 minutes
Correct answer
4 minutes
Why this answer is correct
Given:
Pipe A can empty the tank in 20 minutes and
Pipe B has a diameter twice as that of A
i.e Pipe B has twice the efficiency of Pipe A
Formula Used:
Total Capacity = Efficiency × Time
Concept used:
The efficiency of a pipe is directly proportional to,
The square of the radius of the pipe
Calculation:
The ratio of the diameter of A and B = 1 ∶ 2
The ratio of the radius of pipe A and B = 1/2 ∶ 2/2
The ratio of the radius of pipe A and B = 1 ∶ 2
Now by using the concept,
The ratio of efficiency of A and B = 12 ∶ 22
The ratio of efficiency of A and B = 1 : 4
Time is taken by pipe A to empty the tank = 20 minutes
Total Capacity = Efficiency × Time
⇒ Total capacity of the tank = 20 × 1 = 20 litre
When both Pipe A & B start together,
Then they can empty the whole tank in,
⇒ Total Capacity/Efficiency of both pipe
⇒ 20/(4 + 1)
⇒ 20/5
⇒ 4
∴ Time Required to empty the tank is 4 min.
Alternate MethodPipe A circle area = (π D/2)2
Pipe B circle area = (π 2D/2)2
The efficiency of pipes A and B
⇒ (π D/2)2 = (π 2D/2)2
⇒ 1 : 4
Total capacity of tank = 20 × 1 = 20 litres
When both Pipe A & B start together,
Then they can empty the whole tank in,
⇒ Total Capacity/Efficiency of both pipe
⇒ 20/(4 + 1)
⇒ 20/5
⇒ 4
∴ The time required to empty the tank is 4 min.
Find the value of \(5\frac{1}{3} \div [2\frac{1}{5}\left( {1 - \frac{1}{2}} \right)]\)
160/33
\(4\frac{{26}}{{57}}\)
138/33
None of these
Correct answer
160/33
Why this answer is correct
Concept Used:
Calculation:
First, let us convert all the entities into simple fractions.
\(5\frac{1}{3} ÷ [2\frac{1}{5}\left( {1 - \frac{1}{2}} \right)]\)
\(\begin{array}{l} 5\frac{1}{3} = \frac{{16}}{3}\ 2\frac{1}{5} = \frac{{11}}{5} \end{array}\)
Simplified equation:
(16/3) ÷ (11/10)
⇒ 16/3 × 10/11
⇒ 160/33
A dishonest shopkeeper professes to sell his groceries at his cost price, but uses a false weight of 900 grams for each kilogram. Find his gain percentage.
91/9%
100/9%
100/11%
95/9%
Correct answer
100/9%
Why this answer is correct
Shortcut Trick
For a dishonest dealer selling at cost price, the gain is calculated directly on the false weight given:
Gain % = [Error ÷ False Weight] × 100
⇒ Error = 1000 g − 900 g = 100 g
⇒ Gain % = (100 ÷ 900) × 100
⇒ Gain % = 100/9%
∴ The correct answer is 100/9%.
Alternate Method
Given:
True weight claimed = 1 kg = 1000 grams
False weight used by shopkeeper = 900 grams
Billed Weight (Selling Price) = 1000 g Actual Weight (Cost Price) = 900 g 100g Profit
Calculations:
Let the Cost Price (CP) of 1 gram of groceries be &rupee;1.
⇒ Total CP of the actual goods given (900 grams) = 900 × 1 = &rupee;900
The shopkeeper professes to sell at the CP of 1 kg (1000 grams), so he charges the customer for 1000 grams.
⇒ Selling Price (SP) charged for the 900 grams = 1000 × 1 = &rupee;1000
⇒ Gain = SP − CP = 1000 − 900 = &rupee;100
⇒ Gain % = (Gain ÷ CP) × 100
⇒ Gain % = (100 ÷ 900) × 100
⇒ Gain % = 100/9%
∴ The correct answer is 100/9%.
Additional Information
Dishonest Dealer Concept
When a dealer sells goods at Cost Price but uses a false weight, the profit is made entirely on the weight stolen. The profit percentage must always be calculated on the actual weight delivered to the customer (False Weight). Profit % = [Error ÷ (True Weight − Error)] × 100.
Successive Changes
If a shopkeeper cheats by x% while buying and y% while selling, his overall estimated profit percentage can be calculated using the successive formula: (x + y + x×y ÷ 100)%.
Markup and Discount Relation
If an item is artificially marked up by M% above the Cost Price and a discount of D% is later offered, the net Profit or Loss percentage is: (M − D − M×D ÷ 100)%.
There is a wooden cube with an edge length of 14 cm. A hemisphere with a radius of 7 cm is carved out from each face of the cube. Find the surface area (in cm²) of the remaining structure.\( (Use\pi = \frac{22}{7})\)
1950
2150
1800
2100
Correct answer
2100
Why this answer is correct
Shortcut Trick
Each hemispherical carving removes a circle (πr2) but adds a curved surface (2πr2).
Net surface area gain per face = 2πr2 − πr2 = +πr2.
Total surface area = 6 × (Side2 + πr2).
Total surface area = 6 × (142 + (22/7) × 72).
Total surface area = 6 × (196 + 154) = 6 × 350 = 2100 cm2.
∴ Answer = 2100
Alternate Method
Given:
Edge length of wooden cube (a) = 14 cm
Radius of hemisphere (r) = 7 cm
Number of faces = 6
Formula used:
Total Surface Area = TSA of Cube − 6 × (Area of Circle) + 6 × (CSA of Hemisphere)
Surface Area = 6a2 + 6πr2
Edge (a) = 14 cm r = 7 cm Hemisphere carved from each face
Calculations:
Total Surface Area of the cube = 6 × 142
⇒ 6 × 196 = 1176 cm2
⇒ Surface area removed from 6 faces = 6 × πr2 = 6 × (22/7) × 7 × 7 = 924 cm2
⇒ Surface area added by 6 hemispherical cavities = 6 × 2πr2 = 6 × 2 × (22/7) × 7 × 7 = 1848 cm2
⇒ Net Surface Area = 1176 − 924 + 1848
⇒ Net Surface Area = 1176 + 924 = 2100 cm2
∴ The correct answer is 2100.
Additional Information
Total Surface Area of Cube
The total surface area of a cube with edge 'a' is 6a2. The lateral surface area is 4a2.
Hemisphere Formulas
For a hemisphere of radius 'r', Curved Surface Area = 2πr2 and Volume = (2/3)πr3.
Surface Area Change during Carving
When a hemisphere is carved out from a solid, the total surface area increases by the difference between the curved surface added and the base area removed (πr2).
Volume of Remaining Structure
The volume of the remaining structure is the original volume of the cube (a3) minus the volume of all carved shapes (6 × 2/3πr3).
In what ratio should a grocer mix rice worth Rs.30 per kg and Rs.32.5 per kg so that he can make a profit of 10% by selling the mixture at Rs.34.10 per kg?
5:2
4:5
3:8
3:2
Correct answer
3:2
Why this answer is correct
Given:
Price of 1st type rice = Rs. 30 per kg
Price of 2nd type rice = Rs. 32.5 per kg
The selling price of the mixture = Rs. 34.10 per kg
Profit percentage = 10%
Formula used:
Mixture and allegation method.
Cost price = Selling price × 100/ (100+profit percentage)
Calculation:
Cost price of mixture = 34.10 × 100/(100+10)
= 34.10 × 100/110
= Rs. 31
Ratio of quantities
Ratio of quantities = 1.5: 1
= 3: 2
∴ The correct answer is 3: 2.
A hollow spherical shell is made of a metal of density 6 g/cm3. Its internal and external radius are 2 cm and 4 cm, respectively. What is the weight (in kg) of the shell ? (Take π = \(\frac{22}{7}\))
1.408
1.592
2.73
3.28
Correct answer
1.408
Why this answer is correct
Given:
Density of the metal (ρ) = 6 g/cm3
Internal radius (r1) = 2 cm
External radius (r2) = 4 cm
Volume of a hollow sphere = Volume of outer sphere - Volume of inner sphere
Weight = Volume × Density
Formula used:
Volume of a sphere = (4/3) × π × r3
Calculation:
Volume of outer sphere = (4/3) × π × r23
⇒ Volume of outer sphere = (4/3) × π × 43
⇒ Volume of outer sphere = (4/3) × π × 64 cm3
Volume of inner sphere = (4/3) × π × r13
⇒ Volume of inner sphere = (4/3) × π × 23
⇒ Volume of inner sphere = (4/3) × π × 8 cm3
Volume of hollow sphere
= Volume of outer sphere - Volume of inner sphere
⇒ (4/3) × π × 64 - (4/3) × π × 8
⇒ (4/3) × π × (64 - 8)
⇒ (4/3) × 22/7 × 56 = 704/3 cm3
Mass = Volume × Density
⇒ Mass = 704/3 × 6 = 1408 gm = 1.408 kg
∴ The weight of the shell is approximately 1.408 kg.
A got double the profit than B in a business. A and B invested Rs. 3200 and 2800 respectively. Find the ratio of the number of years A and B invested their amount?
5 : 7
7 : 5
5 : 4
7 : 4
Correct answer
7 : 4
Why this answer is correct
Shortcut Trick
Profit Ratio A : B = 2 : 1
Capital Ratio A : B = 3200 : 2800 = 8 : 7
Time Ratio = (Profit / Capital) Ratio = (2/8) : (1/7)
Time Ratio = (1/4) : (1/7) = 7 : 4
∴ The correct answer is 7 : 4.
Alternate Method
Given:
Profit of A (PA) = 2 × Profit of B (PB)
Investment of A (CA) = ₹3200
Investment of B (CB) = ₹2800
Formula Used:
Profit Ratio = (CapitalA × TimeA) : (CapitalB × TimeB)
Partner A Cap: ₹3200 Time: TA Profit: 2 Units
Partner B Cap: ₹2800 Time: TB Profit: 1 Unit
Ratio of Time (TA : TB) = ?
Calculations:
⇒ PA / PB = (CA × TA) / (CB × TB)
⇒ 2 / 1 = (3200 × TA) / (2800 × TB)
⇒ 2 / 1 = (8 × TA) / (7 × TB)
⇒ 2 × 7 = 8 × (TA / TB)
⇒ 14 = 8 × (TA / TB)
⇒ TA / TB = 14 / 8
⇒ TA / TB = 7 / 4
∴ The correct answer is 7 : 4.
Additional Information
Partnership Profit Distribution
In a partnership, the profit is shared among the partners in the ratio of the product of their investments and the time period for which they invested. Profit = Capital × Time.
Ratio of Investment and Time
If the profit ratio and investment ratio are known, the time ratio can be calculated as T1 : T2 = (P1 / C1) : (P2 / C2).
Simple vs Compound Partnership
A simple partnership involves investment for the same duration, while a compound partnership involves different capitals for different time periods.