Given:
80% of the eligible voters cast their votes,
In which 5% were invalid and
10545 votes are 75% of total valid votes
Calculations:
Let Total eligible voters = z
Total casted votes = 80% of total eligible voters
Total casted votes = z × \(\frac{80}{100}\)
Total valid votes = 95% of total casted votes
Total valid votes = \(\frac{95}{100}\) × \(\frac{80}{100}\) × z
According to question,
10545 = 75% of total valid votes
⇒ 10545 = 75% of ( \(\frac{95}{100}\) × \(\frac{80}{100}\) × z)
⇒ 10545 = (\(\frac{75}{100}\)) × (\(\frac{95}{100}\)) × (\(\frac{80}{100}\)) × z
⇒ 10545 × 1000000 = 570,000 × z
z = Eligible voters = 18500
∴ No. of eligible voters is 18,500.
The following table gives the sales of an electronic chip over 5 years. Find the year in which the sales are equal to the average of the sales over the 5 years.
|
Year |
2015 |
2016 |
2017 |
2018 |
2019 |
|
Sales (In thousands of rupees) |
45 |
54 |
57 |
60 |
69 |
Given:
Sales of an electronic chip over 5 years.
Concept used:
Average = \(\frac{Sum\space of \space Observations}{No. \space of\space observations}\)
Calculations:
Average of the sales over five years
⇒ \(\frac{45+54+57+60+69}{5}\)
⇒ 285/5
⇒ 57
Average of sales over five years = sales in year 2017
∴ The year in which sales is equal to average sales is 2017.
If a + 2b = 27 and a3 + 8b3 = 5427, then find the value of 2ab.
Given:
a + 2b = 27
a3 + 8b3 = 5427
Formula:
a3 + b3 = (a + b) [(a + b)² – 3ab]
Calculation:
a3 + 8b3 = (a + 2b) [(a + 2b)2 – 6ab]
⇒ 5427 = 27 [729 – 6ab]
⇒ \(\frac{5427}{27}\) = (729 – 6ab)
⇒ 201 = 729 – 6ab
⇒ 6ab = 729 – 201
⇒ 6ab = 528
⇒ 2ab = \(\frac{528}{3}\)
⇒ 2ab = 176
∴ The value of 2ab = 176.
Two circles having radii 12 cm and 8 cm, respectively, touch each other externally. A common tangent is drawn to these circles which touch the circles at M and N, respectively. What is the length (in cm) of MN?
Given:
The radii of both circles are 3 cm and 7 cm
Both circles touch each other at point A externally
MN is the common tangent of both circles.
Calculations:
According to the question,
Let the centres of circles be P and Q respectively.
Join P to Q and M. Join Q to N. Draw PT ⊥ QN.
Now PT = MN, as they are opposite sides of rectangle PTNM.
⇒ PQ = PA + AQ
⇒ 12 cm + 8 cm
⇒ 20 cm.
And, QT = QN – NT {As, PM = NT = 8 cm}
⇒ QT = 12 cm – 8 cm
⇒ QT = 4 cm
Now,In Right ∆ PQT, we have:
PT = √ {(PQ)2 – (QT)2 }
⇒ PT = √ {(20)2 – (4)2 }
⇒ PT = √ 384
⇒ PT = 8√6
∴ The length of MN = PT = 8√6 cm.
Shortcut Trick Formula Used:
Length of direct common tangent, when 2 circles are touching externally,
⇒ \(\sqrt{{d^2-(R – r)^2}} \)
where D is the distance between 2 centers
R and r are 2 radii of circles
Calculation:
Using the formula, we get
⇒ \(\sqrt{{{{20^2-(12 – 8)^2}}}}\)
⇒ √384 = 8√6
∴ The length of MN = PT = 8√6 cm.
Avi and Bindu can complete a project in four and twelve hours, respectively. Avi begins project at 5 a.m., and they work alternately for one hour each. When will the project be completed?
Given:
Avi completes a project in 4 hours
Bindu completes the project in 12 hours.
Calculations:
Avi’s 1 hour project is 1/4th part of project
Bindu’s 1 hour project is 1/12th part of project.
Therefore, project done by Avi and Bindu in 2 hours
⇒ 1/4 + 1/12 = 4/12 = 1/3
⇒ 1/3rd part of project
Now,If Time taken by both of them to complete \(\frac{1}{3} \)rd part of project = 2 hours
Then, time taken to complete the whole project =3× 2 = 6 hours
Now ,the project completed at 5 a.m. + 6 hours = 11 a.m.
∴ The project gets completed at 11 a.m.
If x + \(\frac{{1}}{{X}}\) = -2√3, what is the value of x5 + \(\frac{{1}}{{X^5}}\) ?
GIVEN:
x + \(\frac{1}{x}\) = – 2√3
FORMULA USED:
(x + \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) + 2 × x × \(\frac{1}{x}\) = x2 + \(\frac{1}{x^2}\) + 2
(x + \(\frac{1}{x}\))3 = x3 + \(\frac{1}{x^3}\) + 3 × x × \(\frac{1}{x}\) × (x + \(\frac{1}{x}\)) = \(x^3 + \frac{1}{x^3}\) + 3 × (- 2√3 )
CALCULATION:
(– 2√3 )2 = x2 + \(\frac{1}{x^2}\) + 2
⇒ 10 = x2 + \(\frac{1}{x^2}\) —-(1)
Now, (– 2√3)3 = x3 + \(\frac{1}{x^3}\) – 6√3
⇒ – 24√3 + 6√3 = x3 + \(\frac{1}{x^3}\)
⇒ – 18√3 = x3 + \(\frac{1}{x^3}\) —-(2)
Now,
⇒ (x5 + \(\frac{1}{x^5}\)) = (x2 + \(\frac{1}{x^2}\))( x3 + \(\frac{1}{x^3}\) ) – (x + \(\frac{1}{x}\) )
⇒ (x5 + \(\frac{1}{x^5}\)) = 10 × –18√3 – (-2√3) = – 178√3.
∴ The value of (x5 + \(\frac{1}{x^5}\)) is -178√3.
Shortcut Trick If x + \(\frac{1}{x}\) = p, then x5 + \(\frac{1}{x^5}\) = (p2 – 2)(p3 – 3p) – p
If x + \(\frac{1}{x}\) = -2√3, then x5 + \(\frac{1}{x^5}\) = {(-2√3)2 – 2}{(-2√3)3 –
3(-2√3)} – (-2√3) = (12 – 2)(-24√3 + 6√3) – (-2√3) =
10 × -18√3 + 2√3 = -180√3 + 2√3 = -178√3.
On reducing the marked price of his goods by ₹28, a shopkeeper gains 20%. If the cost price of the article be ₹560 and it is sold at the marked price, what will be the gain per cent?
GIVEN:
Gain = 20%
Cost price = Rs560
marked price of goods reduced by Rs28
FORMULA USED:
Gain % = \(\frac{gain}{c.p.}\) × 100
Gain = S.P. – C.P. where S.P. is selling price and C.P. is Cost price
CALCULATION :
20 = \(S.P. – 560\over 560\) × 100
⇒ S.P. = Rs672
As it is sold at marked price ,
So, marked price = Rs (672 + 28 ) = Rs700
So, New S.P. = Rs700.
⇒ Gain = (S.P. – C.P. ) = Rs(700 – 560) = Rs140.
⇒ Gain % = gain /C.P. × 100 = 140/560 × 100 = 25.
∴ Gain percent = 25 %.
If the surface area of a sphere is 64 π cm2, then the volume of the sphere is:
GIVEN:
Surface area of a sphere = \(64 \pi cm^2\)
FORMULA USED:
Surface area of a sphere = \(4 \pi r^2\)
Volume of a sphere = \(\frac{4\pi r^3}{3}\)
CALCULATION:
Surface area of a sphere = 64\(\pi\)
⇒ \(4 \pi r^2\) = \(64\pi\)
⇒ \(r^2\) = 16
⇒ r = 4cm
Now, volume = 4/3 \(\pi\)\(r^3\) = 4/3 × \(\pi\)× 4 × 4 × 4 = \(256 \pi\over3\) \(cm^3\).
∴ Volume of the sphere is \(256 \pi\over3\) \(cm^3\).
Simplify (957 + 932)2 – 4 × 957 × 932.
GIVEN:
\(( 957 + 932 )^2\) – 4 × 957 × 932.
FORMULA USED:
BODMAS
CALCULATION :
⇒ \((1889)^2\) – 4 × 957 × 932
⇒ 3568321 – 4 × 957 × 932
⇒ 3568321 – 3567696 = 625
∴ The value is 625.
Shortcut Trick Formula:
(a+b)2 – 4ab = (a-b)2
Calculation:
\(( 957 + 932 )^2\) – 4 × 957 × 932
= (957 – 932) (957 – 932)
= 25 × 25 = 625
If the numerator of a fraction be increased by 50% and its denominator be diminished by 28%, the value of the fraction is \(\frac{{25}}{{36}}\) Find the original fraction.
GIVEN:
When the numerator of a fraction is increased by 50%and the denominator is decreased by 28%, it becomes 25/36.
CALCULATION :
Let the numerator of a fraction be x and the denominator be y.
⇒ The numerator of a fraction when increased by 50 percent = 150x/100 = 3x/2.
⇒ The denominator of a fraction when decreased by 28 percent = 72y/100 = 18y/ 25.
⇒ Fraction = (3x/2) /(18y/25) = 25x/12y.
According to the question , the fraction obtained is 25/36.
⇒ 25x/12y = 25/36
⇒ x/y = 1/3
⇒ The value of original fraction is 1/3.
If the surface area of a sphere is 1386 cm2, then find the radius of the sphere.
GIVEN:
Surface area of a sphere = 1386 \(cm^2\)
FORMULA USED:
Surface area of a sphere =\(4 \pi r^2\) where r is radius of the sphere.
CALCULATION:
Surface area of a sphere = \(4 \pi r^2\) = 1386
⇒ 4 × \(\frac{22}{7}\) × \(r^2\) = 1386 —(value of \(\pi\) is \(\frac{22}{7}\))
⇒ \(r^2\) = 110.25
⇒ \(r^2\) = \(\frac{11025}{100}\)
⇒ r = \(\sqrt\frac{11025}{100}\) = \(\frac{105}{10}\) = 10.5 cm.
∴ The radius of the sphere is 10.5 cm.
A sum of money becomes ₹3,364 at a rate of 16% compounded annually for 2 years. The sum of money is:
GIVEN:
Amount at the end of 2 years = Rs 3,364
Rate = 16% .
FORMULA USED:
A = P × \(( 1+ \frac{R}{100})^t\) where, A = Amount , P = Principal ,R = Rate of interest and T= Time.
CALCULATION:
A = P × \((1 +\frac{R}{100})^t\)
⇒ 3,364 = P × \(( 1+ \frac{16}{100})^2\)
⇒ 3,364 = P × \(29\over25\)× \(29\over25\)
⇒ P = Rs 2500.
∴ The sum is Rs.2500.
In a class, there are 39 students and their average weight is 51 kg. If we include the weight of the teacher, then the average weight becomes 51.2 kg. What is the weight of the teacher?
GIVEN:
39 students having average weight = 51 kg
39 students and teacher combined average weight = 51.2 kg
FORMULA USED :
Average = \(( sum of all the values)\over( number of all values )\)
CALCULATION :
Average of students = \(\frac{sum}{no.of students}\) = 51
⇒ sum/39 = 51
⇒ sum of weight of students = ( 39 × 51) = 1989.
⇒Average of students and teacher including = \(sum \over( no. of students and teachers)\) = 51.2
⇒ sum/40 = 51.2
⇒ sum of weight of students and teacher = 2048.
weight of teacher = (sum of weight of teacher and students – sum of weight of students )
⇒ ( 2048 – 1989) = 59 kg.
∴ The weight of teacher is 59 kg.
The following pie chart shows the different coloured dresses worn by 60 students on a college party. Study the pie chart and answer the question that follows.
The number of student who wore yellow coloured dress (sector which represents 10%) is
GIVEN:
Different coloured dress worn by 60 students ,which is represented by pie graph
CONCEPT USED :
100 % = Total number of students who wore different colour dresses for a college part
100 % = 60
CALCULATION :
The sector of students who wore yellow colour dress = 10%
⇒ Now,
⇒ 100% = 60
⇒ 10 % = 6.
∴ The number of students wearing yellow colour dress = 6.
The following graph shows the data of the production of electric wire (in thousand tons) by three different companies P, Q and R over the years.
What is the ratio of the average production of Company P in the period 2017 – 2019 to the average production of Company Q in the same period ?
GIVEN:
Three companies namely P,Q and R producing electric wires over the years ( 2015 – 2019 ) respectively
FORMULA USED:
Average = \(( sum of all the values )\over ( number of all the values )\)
CALCULATION :
Company P producing electric wires in 2017 = 25
⇒ P in 2018 = 50
⇒ P in 2019 = 40
Company Q producing electric wires in 2017 = 35
⇒ Q in 2018 = 40
⇒ Q in 2019 = 50
Now,
⇒ sum of all the production in (2017-2019) done by P = ( 25 + 50 + 40 ) = 115
⇒average production of company P = 115 ÷ 3 = 115/3.
⇒sum of all the production in ( 2017 -2019 ) done by Q = ( 35 + 40 + 50 ) = 125
⇒average production of company Q = 125 ÷ 3 = 125/3.
⇒ Ratio of average production done by P and Q = 115/3 : 125/3 = 115 :125 = 23 : 25
∴ Ratio of average production done by P and Q is 23 : 25.
At a certain rate of interest compounded annually, a sum amounts to ₹10,890 in 2 years and to ₹11,979 in 3 years. The sum is:
GIVEN:
Sum amounts to Rs10,890 in 2 years
Sum amounts to Rs11,979 in 3 years
FORMULA USED:
Amount = \(P (1 +\frac{R}{100})^t\) where P = Principal , R = Rate of interest and t = time.
CALCULATION :
10,890 = P× \(( 1+\frac{R}{100})^2\) ……………(1)
11,979 = P × \((1 + \frac{R}{100})^3\) …………….(2)
On dividing Eq (2) by Eq(1) we get,
⇒ 1.1 = \(( 1 + \frac{R}{100})\)
⇒ 0.1 = R/100 = 10
⇒ R = 10 p.a.
Now put the value of R in eq1 we get,
⇒ 10,890 = P× \((1.1)^2\)
⇒ P = 9000 .
⇒ The sum is Rs9000.
Which of the following is divisible by 3 ?
GIVEN :
Numbers divisible of 3 .
CONCEPT USED :
A sum is divisible by 3 when either its sum is divisible by 3 or when its a multiple of 3.
CALCULATION:
Considering option( 1) , we have
⇒7345932
⇒ 7 + 3 + 4 + 5 +9 + 3 + 2 = 33
⇒ 33 ÷ 11 = 3
∴ sum of its digit is divisible by 3 hence, 7345932 is divisible by 3
Considering option (2), we have
⇒ 5439763
⇒ 5 + 4 + 3 +9 +7 + 6 + 3 = 37
⇒ since 37 is not divisible by 3
Hence, the number 5439763 is also not divisible by 3.
Considering option (3), we have
⇒ 3642589
⇒ 3 + 6 + 4 + 2 + 5 + 8 + 9 = 37
⇒since, 37 is not divisible by 3
Hence , the number 3642589 is also not divisible by 3
Considering option (4) ,we have
⇒ 3262735
⇒ 3 + 2 + 6 + 2 + 7 + 3 + 5 = 28
⇒since, 28 is not divisible by 3
Hence, the number 3262735 is also not divisible by 3.
∴ The number 7345932 is divisible by 3 .
A thief was spotted by a policeman from a distance of 225 metres. When the policeman started the chase, the thief also started running. If the speed of the thief was 11 km/h and that of the policeman was 13 km/h, how far would the thief have run, before the policeman caught up with him ?
GIVEN:
At a distance of 225 m policeman spotted a thief
speed of the thief is 11km/h
speed of policeman is 13 km/h :
CONCEPT USED:
Relative speed when the thief and policeman are running in the same direction = ( speed of policeman – speed of the thief)
Distance = Speed × Time
CALCULATION :
Relative speed = ( 13 – 11 ) = 2 km/h
To convert km/h into m/s we have to multiply it by 5/18.
⇒ 2 × 5/18 = 5/9 m/s.
\(Time = \frac{Distance}{Speed}\)
⇒ Time = \(\frac{225}{(5/9)}\) = 225 × \(\frac{9}{5}\) = 405 seconds.
The distance thief had run before he was caught by the policeman
⇒ 11× \(\frac{5}{18}\)× 405 = 1237.5 m
∴ The distance thief had run before he was caught by the policeman is 1237.5 m
A shopkeeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of the cost price to the printed price is:
GIVEN:
Profit = 12%
Discount = 10% on the printed price
FORMULA USED:
Profit% = \(\frac{profit}{c.p.}\)× 100 ,where C.P. and S,P. are Cost price and Selling price respectively
Profit = ( S.P. -C.P. )
Discount = ( printed price – S.P.)
Discount% = \(\frac{( printed price – s.p.)}{printed price}\) × 100
CALCULATION :
Let the cost price be Rs100 ,
⇒ 12 = \(\frac{(S.P. – 100 )}{100}\) × 100
⇒ S.P. = Rs112
Now, let the printed price be x
⇒ 10 = \(\frac{(x-112)}{x}\) × 100
⇒ \(x = \frac{1120}{9}\)
Now, Cost price: Printed price = 100 : \(\frac{1120}{9}\)
⇒ 900 :1120 = 45 : 56.
∴ The ratio of Cost price and Printed price is 45 : 56.
Alternate Method Cost price = (100 – discount%)
Printed price = (100 + profit%)
According to question,
⇒ Cost price = (100 – 10) = 90.
⇒ Printed price = ( 100 + 12) = 112.
⇒ Cost price : Printed price = 90 : 112 = 45 : 56.
∴ The ratio of Cost price and Printed price is 45 : 56 .
In a government scheme, if an electricity bill is paid before the due date, one gets a reduction of 5% on the amount of the bill. By paying the bill before the due date, a person got a reduction of ₹20. The amount of his electricity bill was:
GIVEN:
On paying the bill before the due date , there is 5% reduction on amount of bill
Person got reduction of Rs20
CALCULATION:
Let the amount of the bill be x
Then, 5% of reduction = 0.05x
According to the given question,
⇒ 0.05x = 20
⇒ x = 20/0.05
⇒ 400
∴ The amount of his electricity bill was Rs400.
If cot 75° = 2 – \(\sqrt 3 \). Find the value of cot 15°
GIVEN:
\(cot 75^\circ =2-\sqrt3\)
FORMULA USED :
cot (\(90^\circ-\theta\)) = \(tan\theta\)
\(\frac{1}{tan\theta}\) = \(cot\theta\)
CALCULATION:
\(cot(90^\circ – 15^\circ)\) = \(2-\sqrt3\)
⇒ \(tan15^\circ\) = \(( 2-\sqrt3)\)
⇒ \(cot15^\circ = \frac{1}{tan15^\circ}\) = \(\frac{1}{(2-\sqrt3)}\)
⇒ Now, \(cot15^\circ =\frac {1} {(2-\sqrt3)}\)
On rationalization we get,
⇒ \(cot15^\circ = \frac{1}{(2-\sqrt3)}\)× \(\frac{(2+\sqrt3)}{(2+\sqrt3)}\)
⇒ \((2+\sqrt3)\)/(4-3)
⇒ \((2+\sqrt3)\)/1
⇒The value of \(cot15^\circ \) is \((2+\sqrt3)\).
5 kg of ₹18 per kg wheat is mixed with 2 kg of another type of wheat to get a mixture costing ₹20 per kg. Find the price (per kg) of the costlier wheat.
GIVEN:
Amount of wheat is 5 kg and price is Rs18/kg
Amount of wheat is 2kg
Amount of wheat is 7 kg and price is Rs20/kg
FORMULA USED:
Quantity in kg × amount per kg = Price in Rs.
CALCULATION :
Let the price of 2 kg wheat be Rs. y/kg then,
5 × 18 + 2 × y = 7 × 20
⇒ 90 + 2y = 140
⇒ 2y = 50
⇒ y = 25
∴ The price of costlier wheat is Rs25.
To pack a set of books, Gautam got cartons of a certain height that were 48 inches long and 27 inches wide. If the volume of such a carton was 22.5 cubic feet, what was the height of each carton? [Use 1 foot = 12 inches.]
GIVEN:
Cartons having length = 48 inches and breadth = 27 inches
volume of cartoon = 22.5 cubic feet.
FORMULA USED :
Volume of Cuboid = Length × Breadth × Height
CALCULATION :
Volume of carton = volume of cuboid = Length × Breadth × Height
⇒ volume of carton = 48 × 27 × Height
∵ 1 foot = 12 inches, then 22.5 cubic feet = 22.5 × 12 × 12 ×12
⇒ 22.5 × 12 × 12 × 12 = 48 × 27 × Height
⇒ 38,880 = 1,296 × Height
⇒ Height = 30 inches .
∴ The height of each cartoon is 30 inches.
The fourth proportional to the numbers 5, 6 and 8 is:
GIVEN:
numbers 5,6 and 8
FORMULA USED:
If a, b, c and d are in proportional
then , a/b =c/d.
CALCULATION:
Let fourth number be x
⇒ 5/6 = 8/x
⇒ 5 × x = 6 × 8
⇒ x = 48/5
⇒ x = 9.6.
∴ The fourth proportional is 9.6.
The value of
(43 + 4) ÷ [52 – (72 – 41)] is:
GIVEN:
\((4^3+4)\) ÷ [\(5^2 – (7^2 -41)\) ]
FORMULA USED :
CALCULATION :
⇒ \((4^3 +4 )\) ÷ [\(5^2 – (7^2 -41)\)]
⇒ (64 + 4) ÷ [ 25- 8]
⇒ 68 ÷ 17
⇒ 4.
∴ The value is 4.
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