If x + y + z = 0, then find the value of x3 + y3 + z3 – 3xyz.
Given:
x + y + z = 0
Concept Used:
x3 + y3 + z3 = {(x + y + z) × (x2 + y2 + z2 – xy – yz – zx)} + 3xyz
Calculation:
⇒ x3 + y3 + z3 = {(x + y + z) × (x2 + y2 + z2 – xy – yz – zx)} + 3xyz
By putting the value of (x + y + z = 0),
⇒ x3 + y3 + z3 = {(0) × (x2 + y2 + z2 – xy – yz – zx)} + 3xyz
⇒ x3 + y3 + z3 = 0 + 3xyz
⇒ x3 + y3 + z3 – 3xyz = 0
∴ The value of x3 + y3 + z3 – 3xyz is 0.
If 49 litres of a milk solution has 63% milk in it, then how much milk should be added to make the concentration of milk 70% in the solution?
Given:
Amount of milk solution = 49 litres
Percentage of milk in the solution = 63%
Required concentration of milk in the solution = 70%
Concept Used:
Percentage Composition = Percentage × Solution
Calculation:
Litres of milk in the solution = 63% × 49
Litres of milk in the solution = \(\frac{63}{100}\times49=30\frac{87}{100}\) litres
Litres of water in the solution = 49 – \(30\frac{87}{100}\) = \(\frac{4900-3087}{100}=18\frac{13}{100}\) litres
Now, a solution having 70% milk and 30% water in it,
⇒ If 30% water = \(18\frac{13}{100}\) litres
Then 1% water contains = \(\frac{1813}{100\times 30}=\frac{1813}{3000}\) litres
Then 70% of milk contains = \(\frac{1813}{3000}\times70=\frac{126910}{3000}\) litres
⇒ Amount of milk required to add = \(\frac{126910}{3000}-\frac{3087}{100}\) = \(\frac{126910-92610}{3000}\)
⇒ Amount of milk required to add = \(\frac{34300}{3000}=11\frac{13}{30}\) litres
∴ The amount of milk to be added to the solution to make the concentration of milk 70% is \(11\frac{13}{30}\) litres.
The time taken by a 180 m long train running at a speed of 54 km/h to cross a man standing on the platform is:
Given:
Length of train = 180 m
Speed of train = 54 km/h
Concept Used:
Speed = \(\frac{Distance}{Time}\)
Calculation:
Firstly, we will change the value of speed from km/h to m/s.
⇒ 54 km/h = \(54\times \frac{5}{18}\)
⇒ 54 km/h = 15 m/s
Putting the value in the speed formula, we get
⇒ 15 m/s = \(\frac{180}{Time}\)
⇒ Time = \(\frac{180}{15}\)
⇒ Time = 12 seconds
∴ The time taken by the train to cross a man standing on the platform is 12 seconds.
A worker’s average income for a week is ₹180. He earned ₹115, ₹212, ₹170, ₹165, ₹185, and ₹225 from Monday to Saturday. Find his Sunday’s income.
Given:
Average income for a week = Rs 180
Earning on Monday = Rs 115
Earning on Tuesday = Rs 212
Earning on Wednesday = Rs 170
Earning on Thursday = Rs 165
Earning on Friday = Rs 185
Earning on Saturday = Rs 225
Concept Used:
Average = Sum of total observations / Total number of observations
Calculation:
⇒ Let the earning on Sunday be Rs x.
⇒ Sum of total observations = Rs 115 + 212 + 170 + 165 + 185 + 225 + x
⇒ Total number of observations = 7
⇒ Average = \(\frac{115+212+170+165+185+225+x}{7}\)
⇒ \(\frac{115+212+170+165+185+225+x}{7}=180\)
⇒ 1072 + x = 180 × 7
⇒ 1072 + x = 1260
⇒ x = 1260 – 1072 = Rs 188
∴ The Sunday's income is Rs 188
Find the surface area of a cuboid 17 m long, 8 m broad and 3 m high.
Given:
Length = 17m
Breadth = 8m
Height = 3m
Concept Used:
Surface area of cuboid = 2(lb + bh + hl)
Calculation:
Surface area of cuboid = 2(lb + bh + hl)
⇒ Putting the values in the formula, we get
⇒ 2(17 × 8 + 8 × 3 + 3 × 17)
⇒ 2(136 + 24 + 51)
⇒ 2 × 211 = 422 m2.
∴ The value of surface area of cuboid is 422 m2.
If the total surface area of a cube is 96 cm2, then find the volume of the cube.
Given:
Surface area of cube = 96 cm2
Concept Used:
Surface area of cube = 6a2
Volume of cube = a3
Calculation:
Surface area of cube = 6a2
⇒ Putting the given value of surface area, we get
⇒ 6a2 = 96
⇒ a2 = \(\frac{96}{6}\) = 16
⇒ a = \(\sqrt{16}\) = 4 cm
Side of cube, a = 4 cm
Volume of cube = a3
⇒ a3 = 43
⇒ 64 cm3
∴ The volume of the cube is 64cm3.
The given table shows the consumption of grains (in kilograms) by a village per day over the given set of years. Study the table and answer the following question.
For which set of years (in options) is the combined consumption per day equal to that for the other set of years?
| Year | Rice | Wheat | Maize | Barley |
| 1998-2000 | 181 | 50 | 35 | 26 |
| 2003-2005 | 167 | 44 | 39 | 17 |
| 2008-2010 | 169 | 38 | 49 | 25 |
| 2013-2015 | 150 | 45 | 51 | 21 |
| 2018-2020 | 147 | 54 | 63 | 24 |
Given:
| Year | Rice | Wheat | Maize | Barley |
| 1998-2000 | 181 | 50 | 35 | 26 |
| 2003-2005 | 167 | 44 | 39 | 17 |
| 2008-2010 | 169 | 38 | 49 | 25 |
| 2013-2015 | 150 | 45 | 51 | 21 |
| 2018-2020 | 147 | 54 | 63 | 24 |
Concept Used:
Total consumption per day = Sum of all the weights of each item in kilograms
Calculation:
⇒ Total consumption per day in year 1998-2000 = 181 + 50 + 35 + 26 = 292 kg
⇒ Total consumption per day in year 2003-2005 = 167 + 44 + 39 + 17 = 267 kg
⇒ Total consumption per day in year 2008-2010 = 169 + 38 + 49 + 25 = 281 kg
⇒ Total consumption per day in year 2013-2015 = 150 + 45 + 51 + 21 = 267 kg
⇒ Total consumption per day in year 2018-2020 = 147 + 54 + 63 + 24 = 288 kg
∴ The combined consumption per day is equal for the years 2003-2005 and 2013-2015.
RA and RB are two tangents from a common point ‘R’ outside the circle with centre ‘O’, having a radius of 6 cm. If ∠ARB = 70°, find the measure of the angle AOB.
Given:
Radius = 6 cm
∠ARB = 70°
Concept Used:
Sum of the angles of a quadrilateral is 360°
The angle between the radius and the tangent of a circle is 90°
Calculation:
⇒ The 4 points A, B, R and O make a quadrilateral
⇒ ∠RAO = ∠RBO = 90°
⇒ ∠RAO + ∠AOB + ∠RBO + ∠ARB = 360°
⇒ 90° + ∠AOB + 90° + 70° = 360°
⇒ ∠AOB = 360° – 250° = 110°
∴ The value of ∠AOB is 110°.
On dividing 8675123 by a certain number, the quotient is 33611 and the remainder is 3485. The divisor is ______.
Given:
Number = 8675123
Quotient = 33611
Remainder = 3485
Concept Used:
Number = Quotient × Divisor + Remainder
Calculations:
⇒ 8675123 = 33611 × Divisor + 3485
⇒ 8675123 – 3485 = 33611 × Divisor
⇒ 8671638 = 33611 × Divisor
⇒ Divisor = \(\frac{8671638}{33611}\)
⇒ Divisor = 258
∴ The required value of divisor is 258.
A shopkeeper wanted to sell ₹2,700 worth of products. But he had two options — giving two successive discounts of 10% and 15%, respectively, or giving a single discount of 25%. What was the difference between the two options?
Given:
Product Price = Rs 2,700
Successive Discounts = 10% and 15%
Single Discount = 25%
Concept Used:
Discount = % of Discount × Product Price
Price After Discount = Product Price – Discount
Calculations:
For first successive discount of 10%,
⇒ Product Price \(=2,700-10\% \times2,700 \)
⇒ Product Price \(=2,700-\frac{10}{100}\times2,700 \)
⇒ Product Price \(=2700-270 \)
⇒ Product Price = Rs 2,430
For second successive discount of 15%,
⇒ Product Price \(=2430-15\%\times2430 \)
⇒ Product Price \(=2430-\frac{15}{100}\times2430 \)
⇒ Product Price \(=2430-364.50\)
⇒ Product Price = Rs 2065.50
Price after single discount of 25%,
⇒ Product Price \(=2700-25\%\times 2700 \)
⇒ Product Price \(=2700-\frac{25}{100}\times2700 \)
⇒ Product Price \(=2700-675\)
⇒ Product Price = Rs 2025.00
Difference between prices after the discounts = Rs 2065.50 – Rs 2025.00
Difference between prices after the discounts = Rs 40.50
∴ The required difference between the two options is Rs 40.50
In a Panchayat election, three candidates were there in the fray. The losing candidates got an equal number of votes and the winning candidate got as many votes as the rest of the two got collectively. If the total votes polled were 1500, find the number of votes that the winning candidate got, assume all votes polled were valid.
Given:
Winner got the votes equal to the sum of rest two candidates
Rest candidates got equal amount of votes .
Calculations:
Let the loosing candidates got x votes each
Then, winner got 2x votes.
Now, according to question,
2x + x + x = 1500
⇒ x = 375
⇒ Winner got 2x = 2 × 375
∴ Winner got 2x = 750 votes.
The cost price of an article is Rs.6,450. If it sold at a profit 16%, how much would be its selling price?
Given:
Cost price = Rs.6, 450
Profit = 16%
Concept:
Selling price = \(\frac{(100+profit)}{100}×Cost \space price\)
Calculation:
Selling price = \(\frac{(100+16)}{100}×6, 450\)
⇒ (116/100) × 6450
⇒ 58 ×129
∴ The Selling price is Rs. 7482.
16 men can paint 15 walls in 45 days. In how many days can 20 men paint 20 walls?
Given:
16 men can paint 15 walls in 45 days.
Calculations:
To paint 15 walls, 16 men take = 45 days
To paint 1 wall, 16 men will take = \(\frac{45}{15}\) = 3 days
To paint 1 wall, 1 men will take = \(16×3 \) = 48 days
To paint 1 wall, 20 men will take = \(\frac{1}{20}×48 \) = \(\frac{12}{5}\) days
To paint 20 walls, 20 men will take = \(20×\frac{12}{5}\) = 48 days
∴ 20 men will take 48 days to paint 20 walls.
Alternate Method
Given:
M1 = 16, D1 = 45, W1 = 15
M2 = 20, W2 = 20
Concept used:
(M1 × D1)/W1 = (M2 × D2)/W2
Calculations:
(M1 × D1)/W1 = (M2 × D2)/W2
⇒ \(\frac{16 × 45}{15}\) = \(\frac{20 × D2}{20}\)
⇒ 48 = D2
Hence, 20 men will take 48 days to paint 20 walls.
A pond is 50 m long, 35 m wide and 2.5 m deep. Find the capacity of the pond.
Given:
Length = 50 m
Breadth = 35 m
Width = 2.5 m
Concept:
Volume of cuboid = length × breadth × height
Calculations:
Volume of pond =\(50×35×2.5\)
⇒ Volume= 4375 m³
∴ Capacity of the Pond is 4375 m³.
If, in a family, consumption of sugar is 6 kg for 24 days, then what is the consumption of sugar for 18 days?
Given:
Consumption of sugar for 24 days is 6 kg.
Calculations:
If Consumption of sugar for 24 days is 6 kg
⇒ consumption of sugar for 1 days is \(\frac{6}{24}\)kg
⇒ consumption of sugar for 18 days is \(\frac{1}{4}×18=4.5\) kg
∴ The consumption of sugar for 18 days is 4.5 kg.
Simplify [7\(\frac{1}{2}\) ÷ {1\(\frac{1}{4}\) – \(\frac{1}{2}\) (2\(\frac{1}{2}\) – \(\overline {\frac{1}{4} – \frac{1}{6}} \))}] ÷ (\(\frac{1}{2}\) of 8\(\frac{1}{3}\)).
Given:
[7\(\frac{1}{2}\) ÷ {1\(\frac{1}{4}\) – \(\frac{1}{2}\) (2\(\frac{1}{2}\) – \(\overline {\frac{1}{4} – \frac{1}{6}} \))}] ÷ (\(\frac{1}{2}\) of 8\(\frac{1}{3}\))
Concept used:
VBODMAS
Calculations:
[7\(\frac{1}{2}\) ÷ {1\(\frac{1}{4}\) – \(\frac{1}{2}\) (2\(\frac{1}{2}\) – \(\overline {\frac{1}{4} – \frac{1}{6}} \))}] ÷ (\(\frac{1}{2}\) of 8\(\frac{1}{3}\))
⇒ \([7\frac{1}{2}÷[1\frac{1}{4}-\frac{1}{2}(2\frac{1}{2}-\frac{1}{12})]÷(\frac{1}{2}of8\frac{1}{3})\)
⇒ \([\frac{15}{2}÷[\frac{5}{4}-\frac{1}{2}(\frac{5}{2}-\frac{1}{12})]]÷(\frac{1}{2}×\frac{25}{3})\)
⇒ \([\frac{15}{2}÷[\frac{5}{4}-\frac{1}{2}(\frac{29}{12})]]÷(\frac{1}{2}×\frac{25}{3})\)
⇒ \([\frac{15}{2}÷[\frac{5}{4}-\frac{29}{24})]]÷(\frac{25}{6})\)
⇒ \([\frac{15}{2}÷[\frac{1}{24}]]÷(\frac{25}{6})\)
⇒ \([\frac{15}{2}\times[\frac{24}{1}]]÷(\frac{25}{6})\)
⇒ \([15\times[\frac{12}{1}]]÷(\frac{25}{6})\)
⇒ \(180÷(\frac{25}{6})\)
⇒ \(180×\frac{6}{25}\) = 43.2
\(\therefore \)The value required is 43.2.
In a company there are only two types of employees, workers and officers. 35% of the employees are officers and the average monthly salary of an officer is ₹39,000 more than the combined average monthly salary of an employee. What is the difference between the average monthly salaries of an officer and a worker?
Given:
Average monthly salary of an officer is Rs. 39000 more than the average monthly salary of an employee and 35 % of employees are officers and rest are workers.
Concept used:
Sum of Observations = No. of Observations × Average value
Calculations:
Let total no. of employees be 100 and
let the average monthly salary of officers and workers be X and Y respectively.
Then, the no. of officers be 35 and workers be 65.
Now, According to question,
X = Average monthly salary of an employee + 39,000
⇒ X = \(\frac{(35X+65Y)}{100}+39,000\)
⇒ \( 100X= 35X+65Y+3,900,000\)
⇒ \(100X-35X-65Y=3,900,000 \)
⇒ \( 65(X-Y)=3,900,000\)
⇒ \( X-Y=\frac{3,900,000}{65}\) = 60,000
∴ The required difference is Rs. 60,000.
Study the given graph carefully and answer the question that follows.
In which of the following years was the sales of company 1 exactly half of the total sales of companies 2 and 3 together in that year?
Given:
Graph representing sales of three companies in each year.
Calculations:
According to question,
Half of the sum of sales of companies 2 and 3 together = sales of company 1
Now , in year 2015, we have
⇒ \(\frac{(50+40)}{2} = 45\)
Which is not equal to 30 i.e., sales of company 1 in that year.
Similarly, In the year 2016, we have
⇒ \(\frac{(60+40)}{2}\) = 50
Which is exactly equal to 50 i.e., sales of company 1 in that particular year.
\(\therefore\) The year in which half of the sales of 2nd and 3rd company is equal to sales of first company is 2016.
What will be a single equivalent discount for successive discounts of 20%, 40% and 75% on the marked price of an item?
Given:
The successive discount are 20%, 40% and 75%.
Formula:
Successive discount = X + Y \(-\frac{XY}{100}\)
Where,
X = First discount
Y = Second discount
Calculation:
Successive discount = \(X+Y-\frac{XY}{100}\)
\(20+40-\frac{(20×40)}{100}\) = 52%
Now, again
Successive discount = \(X+Y-\frac{XY}{100}\)
⇒ X = 52%, Y = 75%
⇒ \(52+75-\frac{(52×75)}{100}\)
⇒ \(127-39\)
⇒ \(88\)
∴ The single equivalent discount is 88%.
The population of a city increases at the rate of 10% per annum. If the population of the city was 2 crore in 2018, then the population of the city in the year 2021 was:
Given:
Principal = 2 crores
Rate = 10%
Time = 2021 – 2018 = 3 years
Formula used:
Amount = \(P(1+\frac{R}{100})\)T
Calculation:
Amount = \(P(1+\frac{R}{100})\)T
⇒ \(20,000,000(1+\frac{10}{100})³\)
⇒ \(20,000,000(\frac{11}{10})³\)
⇒ \(20,000,000×\frac{1331}{1000}\)
⇒ Amount = 26,620,000
∴ The population in 2021 is 2.662 crores.
If (x + \(\frac{1}{x}\)) = 7.5, what is the value of (x3 + \(\frac{1}{{{x^3}}}\))?
If x + \(\frac{1}{X}\) = – 13, what is the value of x4 + \(\frac{1}{{{X^4}}}\) ?
Given:
x + \(\frac{1}{x}\) = – 13
Formula used:
a² + b² = (a + b)² – 2ab
Calculation:
x2 + \(\frac{1}{x^2}\) = (x + \(\frac{1}{x}\))2 – 2 × x × \(\frac{1}{x}\)
⇒ x2 + \(\frac{1}{x^2}\) = 132 – 2
⇒ x2 + \(\frac{1}{x^2}\) = 167
Now,
x4 + \(\frac{1}{x^4}\) = (x2 + \(\frac{1}{x^2}\))2 – 2
⇒ (167)2 -2
⇒ 27889 – 2 = 27887
⇒ x4 + \(\frac{1}{x^4}\) = 27887
∴ The value of x4 + \(\frac{1}{x^4}\) = 27887.
Shortcut TrickIf x + \(\frac{1}{x}\) = p, then x2 + \(\frac{1}{x^2}\) = p2 – 2
If x + \(\frac{1}{x}\) = – 13, then x2 + \(\frac{1}{x^2}\) = (-13)2 – 2 = 167 = k(Let)
If x2 + \(\frac{1}{x^2}\) = k, then x4 + \(\frac{1}{x^4}\) = k2 – 2
If x2 + \(\frac{1}{x^2}\) = 167, then x4 + \(\frac{1}{x^4}\) = (167)2 – 2 = 27887
If sec θ – cos θ = 14 and 14 sec θ = x, then the value of x is _________.
Given:
secθ – cosθ = 14 and 14 secθ = x
Concept used:
\(Sec\theta =\frac{1}{Cos\theta}\)
Calculations:
According to the question,
⇒ \(sec\theta – cos\theta= 14\)
⇒ \(\sec\theta-\frac{1}{sec\theta}=14\)
⇒ \( sec²\theta-1=14sec\theta\)
⇒ \(\tan^2\theta=14sec\theta\) —-(\(sec²\theta-1=tan^2\theta\))
\(\ tan²\theta=x\)
∴ The value of x is \(tan²\theta\).
The table below provides information about the marks obtained by a group of students in a test.
| Marks |
No. of Students |
| Less than 10 |
2 |
| Less than 20 |
5 |
| Less than 30 |
6 |
| Less than 40 |
8 |
| Less than 50 |
10 |
How many students scored at least 30 marks but less than 40 marks in the test?
Given:
| Marks | No. of students |
| Less than 10 | 2 |
| Less than 20 | 5 |
| Less than 30 | 6 |
| Less than 40 | 8 |
| Less than 50 | 10 |
Calculations:
Students scored at least 30 marks but less than 40 marks on the test
⇒ Students scored less than 40 – Students scored less than 30
⇒ 8 – 6 = 2
∴ Students scored at least 30 marks, but less than 40 marks on the test is 2.
In Class 8 of a school, the ratio of students playing cricket to those playing football is 27 ∶ 13, while the ratio of those playing football to those playing hockey is 26 ∶ 14. If the total number of students in Class 8 is 470, and each student plays exactly one game, find the percentage of students of the class playing cricket. [Give your answer correct to two decimal places.]
Given:
Ratio of students playing cricket to those playing football is 27 : 13 and
Ratio of students playing football to those playing hockey is 26 : 14.
Total no. of students is 470
Calculations:
Ratio of students playing cricket, football and hockey respectively is 27 × 26 : 26 × 13 : 13 × 14
⇒ 702 : 338: 182
⇒ 351: 169 : 91
⇒ 27 : 13 : 7
Now, according to the question,
(27 + 13 + 7)k = 470
⇒ 47 × k = 470
⇒ k = 10
Now , no. of students playing cricket = 27k = 27 × 10 = 270
Percentage required = 270/470 × 100 = 57.45%
∴ The percentage reduired = 57.45%