In a ΔABC, right angled at B, if tan A = \(\frac{1}{{\sqrt 3 }}\), then sin A . cos C + cos A . sin C = ______.
GIVEN:
Triangle ABC ,right angled at B i.e. angle B = \(90^\circ\)
tan A = \(1\over\sqrt3\)
FORMULA USED:
tan A = P/B
sin A= P/H
Cos C= B/H where , P = perpendicular , B = base and H = hypotenuse
Tan 30° = \(1\over \sqrt3\) = \(P \over B\)
\(H^2 = P^2 + B^2\)
CALCULATION:
tan A = \(1\over \sqrt3\) = \(P\over B\)
⇒ \( H ^2 = P^2 +B^2\)
⇒ \(( 1^2 + \sqrt3^2)\) = 4
⇒ \(H^2 = 4\)
H = 2 unit .
Now ,
sin A = P/H = 1/2 unit .
⇒ Cos A = B/H = \(\sqrt 3\) /2 unit,
⇒ sin C = P/H = \(\sqrt3\) / 2 unit .
⇒ Cos C = B/H = 1/2 unit .
Hence, sin A .Cos C + Cos A . sin C = 1/2 ×1/2 + \(\sqrt3\over2\) × \(\sqrt3\over2\)
⇒ (1/4+ 3/4 )
⇒ 4/4
⇒1
∴ sin A . cos C + cos A . sin C = 1.
Alternate Method tan A = \(1\over\sqrt3\)
⇒ A = \(30^\circ\)
As, B = \(90^\circ\)
In a triangle of ABC , the sum of angle A + angle B + angle C = \(180^\circ\)
⇒ \(( 30^\circ +90 ^\circ)\) + angle C = \(180^\circ\)
⇒ angle C = \(60^\circ\) .
⇒ Now, sin A = sin 30 = 1/2
⇒ sin C = sin 60 = \(\sqrt3\) /2
⇒ Cos A = COS 30 = \(\sqrt3\) /2
⇒ Cos C = COS60 = 1/2
∴ Sin A .Cos C + Cos A + sin C = 1/2 × 1/2 + \(\sqrt3\) /2 ×\(\sqrt3\) /2 = 1.
The population of a town is 80,000. It decreases by 8% in the first year and increases by 5% in the second year. What will be the population of the town at the end of 2 years?
GIVEN:
Population = 80,000
population decreases in the 1st year = 8%
population increases in the 2nd year = 5%
FORMULA USED:
New value = ( 100 + X)/100 × Actual value , where X = percent by which value is increased.
New value = ( 100 – X)/100 × Actual value , where X = percent by which value is decreased.
CALCULATION:
Population of town after 1st year = 80,000 – 80,000 × 8 /100 = (80,000- 6,400) = 73,600
Population of town after 2nd year = 73,600 + 73,600 × 5/100 = ( 73,600 + 3,680) = 77,280
∴ The population at the end of 2 years is 77,280.
If Mona and Sona together can finish a typing work in 18 days and Sona alone can finish it in 24 days, find the number of days in which Mona alone can finish the work.
GIVEN:
Mona and Sona can finish typing in 18 days
sona alone can finish in 24 days
CONCEPT USED:
No. of days = total work /efficiency
CALCULATION :
Let the total amount work be 72 unit —(LCM of 18 and 24)
⇒ mona and sona finish 72 unit work in 18 days
⇒ ( mona + sona )’s efficiency = total work /no. of days = 72 ÷ 18 = 4 —-(1)
⇔ sona alone finishes 72 unit work in 24 days
efficiency of sona = total work ÷ no.of days = 72 ÷ 24 = 3 —-(2)
Now, on subtracting eq.(1) from eq (2), we get
mona efficiency = ( mona + sona )’s efficiency – sona efficiency = (4 -3) =1
⇒ No. of days in which mona alone finishes work = total work ÷ mona’s efficiency
⇒ ( 72 ÷ 1 ) = 72 days .
∴ No. of days in which mona finishes work is 72 days .
X’s salary is increased by 20% and then decreased by 20%. What is the change in salary?
GIVEN:
X’s salary is increased by 20% and then decreased by 20% .
FORMULA USED:
when one percentage change is positive and the other is negative ,
net percentage change = (x-y-xy/100)% , where x and y are successive percentage change respectively.
CALCULATION:
When X’s salary is increased i.e. the percentage change is positive = 20%
When X’s salary is decreased i.e. the percentage change is negative = 20%
Hence, net percentage change = ( 20 -20 -400/100)% = -4% , where – sign shows decrease .
∴ Change in salary is -4% i.e. 4% decrease .
Alternate Method
FORMULA USED :
Change in salary = ( new salary -original salary )/original salary × 100
Let the initial salary of X’s be 100
20% increase in salary = 100 + (20% of 100) = 100+20 = 120.
20% decrease in salary = 120 – (20% of 120) = 120- 24 = 96.
Hence, new salary after increment and decrement = 96
original salary = 100
decrease in salary = (new salary – original salary)/original salary × 100
⇒ \((96 -100)\over 100\) × 100 = -4% where , – sign shows decrease
∴ Overall change in salary is -4% or decrease in 4%.
The given circle graph shows the spending of a country on various sports during a year. Study the graph carefully and answer the question that follows.
The country spent the same amount of money on:
GIVEN:
spending of a country on various sports.
Others – 4% , football – 15%, hockey -10%, golf- 10%, tennis-6%,cricket-30%, basketball-25%
CALCULATION:
Let the total amount spend on various sports be 100
Now,
Considering option(1), we have
⇒ amount spent on cricket = 30% of 100 = 30
⇒ amount spent on football = 15% of football = 15
Hence, the amount spent on cricket and football are not same
Considering option (2), we have
⇒ amount spent on golf = 10% of 100 = 10
⇒ amount spent on basketball = 25% of 100 = 25
Hence, the amount spent on golf and basketball are not same.
Considering option (3) , we have
⇒ amount spent on hockey = 10% of 100 = 10
⇒ amount spent on tennis = 6% of 100 =6
Hence, the amount spent on hockey and tennis are not same .
Considering option (4), we have
⇒ amount spent on hockey = 10% of 100 =10
⇒ amount spent on golf = 10% of 100 = 10.
∴ The amount spent on Hockey and Golf are same.
If 2a + 5b = 12 and ab = 3, find the value of 4a2 + 25b2.
Given:
2a+5b = 12, ab = 3
Formula used:
\((a+b)^2=a^2+b^2+2ab\)
Calculation:
Take the square of the expression, (2a+5b)
\((2a+5b)^2=(2a)^2+(5b)^2+2(2a)(5b)\)
\((2a+5b)^2=4a^2+25b^2+20(ab)\)
It is given that (2a + 5b) = 12 and ab = 3
On substituting the values in the expression
\((12)^2=4a^2+25b^2+20 \times (3)\)
\(144-60=4a^2+25b^2\)
\(4a^2+25b^2=84\)
∴ The value of \(4a^2+25b^2\) is 84.
The difference between the compound interest and the simple interest on a given principal is ₹1,725 in 2 years and the rate of interest in both cases is 25% per annum, and in the case when interest is compunded, it is compounded annually. Find the principal.
Given:-
Rate = 25% ,
Time period = 2,
CI – SI = Rs.1725
To Find:-
Principal amount
Formula used:-
SI = P × R × T and CI = \(P(1+\frac{R}{n})^{nT}-P\)
Calculation:
It is given that the compound interest is compounded annually which means n = 1
⇒CI = \(P(1+R)^{T}-P\)
As mentioned in the question,
CI – SI = 1725
\(P(1+R)^{T}-P\) – P × R × T = 1725
On substituting the values of R and T in the given equation by 25% and 2 respectively then,
\(P\left(1+\frac{25}{100}\right)^2-P – \left(P\times\frac{25}{100}\times 2\right)=1725\)
On solving,
\(P\left(\frac{5}{4}\right)^2-P-\left(\frac{P}{2}\right)=1725\)
\(\frac{25P}{16}-\frac{3P}{2}= 1725\)
P = Rs.27600
The principal amount is Rs. 27600.
What is the volume of the largest right circular cone that can be cut out from a cube whose edge is 10 cm?
Given:
Edge of cube(a) = 10cm
Concept Used:
Height of cone(h) = edge of cube and, radius of cone(r) = \(\frac{1}{2}\)(edge of cube)
Volume of cone = \(\frac{1}{3}\pi r^2h\)
Calculation:
Let the height be ‘h’ cm and radius be ‘r’ cm
For volume to be maximum,
⇒h = a and r = \(\frac{a}{2}\)
Volume = \(\frac{1}{3}\pi r^2h\) —-(i)
Substitute the value of ‘h‘ and ‘r‘ into the expression (i),then
Volume = \(\frac{1}{3}\pi {(\frac{a}{2})}^2(a)\)
On putting the value of a,
Volume = \(\frac{1}{3}\pi {(\frac{10}{2})}^2 (10)\)
Volume = \(\frac{1}{3}\pi \times 25 \times 10\)
Volume = \(\frac{250\pi}{3}\)
The Maximum volume of right circular cone is \(\frac{250\pi}{3} \ cm^3\)
The price of an article is raised by 45% and then two successive discounts of 15% each are allowed. Ultimately the price of the article is ______.
Given:
Price raised by 45% and two successive discount of 15% is given
Formula used:
Percent change = \(\frac{final\ value-initial\ value}{initial\ value}\times 100\)
Calculation:
Let the cost price of an article is Rs100
As mentioned, the price is raised by 45% that means
New price = Old price + 45% of old price
⇒ 100 + \(\frac{45}{100}\times 100\)
New price = 145 rs
Since two successive discount of 15% has been given on new price,
After first discount,
Discounted Price =145 – 15% of 145
Discounted Price =145 – \(\frac{15}{100}\times145\)
Discounted Price =123.25 rs
After second discount,
Final price = 123.25 – 15% of 123.25
123.25 – \(\frac{15}{100}\times 123.25\)
123.25 – 18.4875
Final price = 104.7625
Percentage change in price = \(\frac{final\ value-initial\ value}{initial\ value}\times 100\)
% change = \(\frac{104.7625-100}{100}\times 100\)
%change = 4.7625 %
The price of the article is increased by 4.7625 %.
How many two-digit numbers are divisible by 3 or 5?
Given:
Divisible by 3 or 5
Concept used:
Using the formula of finding the nth term of an AP
an = a + (n – 1) × d
Calculation:
The first two-digit number divisible by 3 is 12
The second two-digit number divisible by 3 is 15
The third is 18
We can see that an AP is formed here, whose first term is 12 and the common difference is 3
By hit and trial we can find the last two-digit number divisible by 3 will 99, which means the last term of an AP is 99
Here, a = 12 , d = 3 and an = 99
By putting the values in the formula,
99 = 12 + (n – 1) × 3
n – 1 = 29
n = 30
The two-digit numbers divisible by 3 are 30
Similarly, by taking 10 as the first and 95 as the last digit we could find the number of two digit numbers divisible by 5,
Here, a = 10 , d = 5 and an = 95
By putting the values in the formula,
95 = 10 + (n – 1) × 5
n – 1 = 17
n = 18
The two-digit numbers divisible by 5 is 18
It is asked in the question that the two-digit number should either be divisible by 3 or 5 but not by both,
The number of two-digit numbers divisible by both can be found by forming AP(Arithmetic progression) as same as above-
Here, the first term and the common difference will be L.C.M.(3 , 5) i.e. 15
⇒By hit and trial, the last term will be 90
Using nth term formula of an A.P , an = a + (n +1) × d
⇒90 = 15 + (n – 1) × 15
75 = (n – 1) × 15
5 = (n – 1)
.·. n = 6
The two-digit numbers divisible by both 3 and 5 are 6.
So, The total number of two-digit numbers divisible by 3 or 5 will be,
(Number divisible by 3) + (Number divisible by 5) – (Number divisible by both 3 and 5)
⇒30 + 18 – 6 = 42
∴ The two-digit numbers divisible by 3 or 5 are 42.
The total surface area of a solid hemisphere is 16632 cm2. Its volume is: (Take π = 22/7)
Given:
TSA of Hemisphere = \(16632 cm^2\)
Formula used:
Volume of solid hemisphere = \(\frac{2}{3}\pi r^3\)
Calculation:-
It is given that the total surface area of the hemisphere is \(16632 cm^2\)
TSA(Total surface area) = \(3\pi r^ 2\)
\(16632=3\pi r^2\)
\(r^2=\frac{5544}{\pi}\)
\(r^2=1764\)
r = 42 cm
Volume = \(\frac{2}{3}\pi r^3\)
V = \(\frac{2}{3}\times \frac{22}{7} \times r^3\) = \(\frac{2}{3}\times \frac{22}{7} \times 42^3\)
On solving,
V = 155232 cm3
∴ The volume of solid hemisphere is 155232 cm3.
If P : Q = 10 : 11 and Q : R = 11 : 12, then P + Q : Q + R : R + P is:
Given:
P : Q = 10 : 11 and Q : R = 11 : 12
Concept used:
Combining ratios
Calculation:
P : Q = 10 : 11 —-(i)
Q : R = 11 : 12 —-(ii)
Since in both the ratios (i) and (ii) the combining one,Q, has the same value in both the cases
Hence, P : Q : R = 10 : 11 : 12
Let P , Q , and R be 10x ,11x , and 12x respectively, where x is the factor by which ratio is changing-
then, P + Q = 10x + 11x = 21x
Q + R = 11x +12x = 23x
P + R = 10x + 12x = 22x
⇒ P + Q : Q + R : R + P = 21x : 23x : 22x
⇒P + Q : Q + R : R + P = 21 : 23 : 22 (by putting x = 1)
∴ The value of P + Q : Q + R : R + P is 21 : 23 : 22
Find the greatest among the following options.
Given:
0.62 , \(\frac{2}{3}\) , 0.57 , \(\frac{4}{5}\)
Concept Used:
Decimal conversion and comparison.
Calculation:
\(\frac{2}{3}\) could be written as 0.67 in decimal form
and,
\(\frac{4}{5}\) could be written as 0.8 in decimal form.
On comparison between 0.62 , \(\frac{2}{3}\) , 0.57 , \(\frac{4}{5}\)
0.8 = \(\frac{4}{5}\) > 0.67 = \(\frac{2}{3}\) > 0.62 > 0.57
∴ \(\frac{4}{5}\) is the greatest among all other options
The average marks scored by a student in 4 subjects is 75. But when the marks of English are added to it, the overall average became 70. How much did he score in English?
Given:
Average marks=75
Formula used:
Average = \(\frac{sum\ of\ observation}{total\ number\ of\ observation}\)
Calculation:
Total marks scored by student in 4 subjects = Average × 4
Total marks of 4 subjects = 75 × 4 = 300
Let the marks in English be ‘x’, then on adding
New Average = 70
New Average = \(\frac{300+x}{5}\)
\(70=\frac{300 + x}{5}\)
⇒ x = 50
∴ The mark scored by student in English is 50.
A dealer professes to sell his goods at cost price, but uses 900 gm weight for a kilogram. Find his profit per cent.
Given:
Selling price = Cost price,
Using 900gm in place of 1000gm
Formula used:
% Profit = \(\frac{Selling\ Price\ -\ Cost\ Price}{Cost\ Price}\times100\)
Calculation:
Let the price of 1gm of the product be ‘x‘
then, the selling price of 1kg(1000gm) of the product will be 1000x
as mentioned in the question that he is using 900gm of weight in place of 1000gm weight which means that the amount of product he will be selling is 900gm
⇒The C.P. of the product will be 900x
By using the formula of profit percentage,
% profit = \(\frac{1000x-900x}{900x}\times 100\)
% profit = \(\frac{100x}{900x}\times 100\)
% profit = \(\frac{100}{9} \%\) = \(11\frac{1}{9}\%\)
∴ The dealer will be getting a profit of \(11\frac{1}{9}\%\) .
If triangles ABC and PQR are both isosceles with AB = AC and PQ = PR, respectively. If also AB = PQ and BC = QR and angle B = 50°, then what is the measure of angle R?
Given:
ABC and PQR are isosceles triangles, AB = AC, PQ = PR and
AB = PQ and BC = QR and \(\angle B=50^{\circ}\)
Concept used:
Congruency criterion of triangles
Calculation:
It is given that AB = AC and PQ = PR but AB = PQ
⇒ AB = AC = PQ = PR ——-(I)
In \(\triangle ABC \) and \(\triangle PQR\)
AB = PQ (By equation-I )
BC = QR (Given)
AC = PR (By equation-I )
Now, By SSS (side-side-side) congruency criterion
\(\triangle ABC \cong \triangle PQR\)
Then, By Corresponding parts of congruent triangles-
\(\angle ABC=\angle PQR= 50^{\circ}\)
Since, \(\triangle PQR\) is an isosceles triangle
\(\angle PQR=\angle PRQ\)
⇒\(\angle PRQ=50^{\circ}\)
.·. The measure of angle R is 50°.
How many 25 cm × 11.25 cm × 6 cm bricks will be required to construct an 8 m × 6 m × 22.5 cm wall? (ignoring other material used)
Given:- Dimension of brick= 25 cm × 11.25 cm × 6 cm
To find:- Number of bricks
Formula used:
Volume of cuboid = Length × width × height
Calculation:
The volume of the brick is
Volume of Brick = 25 cm × 11.25 cm × 6 cm = 1687.5 \(cm^3\)
The volume of the wall( to be constructed) is,
Volume of wall = 8 m × 6 m × 22.5 cm
(Change the values of dimension in ‘cm’ as it is given in ‘m’)
Volume of wall = 800 cm × 600 cm × 22.5 cm —-(1 m = 100 cm )
⇒ \(108 \times 10^5 cm^3\)
Number of brick required = \(\frac{Volume\ of\ wall}{volume\ of\ brick}\)
⇒ \(\frac{108 \times 10^5 \ cm^3}{1687.5\ cm^3}\) = 6400
∴ The number of bricks required is 6400.
If \(\left[ {3\frac{6}{7}\, ÷ \,\frac{{54}}{7}\, – \,\left\{ {3\, – \,\left( {2\frac{3}{4}\, – \,\frac{3}{2}} \right)} \right\}} \right]\, \)+ A ÷ 4 = 0, then what is the value of A?
Given:-\(\left[ {3\frac{6}{7}\, ÷ \,\frac{{54}}{7}\, – \,\left\{ {3\, – \,\left( {2\frac{3}{4}\, – \,\frac{3}{2}} \right)} \right\}} \right]\, \)+A ÷ 4 = 0
To find:- Value of A
Concept used:- BODMAS
Calculation:
The expression could be written as-
\(\left[ {3\frac{6}{7}\, ÷ \,\frac{{54}}{7}\, – \,\left\{ {3\, – \,\left( {2\frac{3}{4}\, – \,\frac{3}{2}} \right)} \right\}} \right]+\frac{A}{4}=0\)
Solving the expression inside () small bracket,
\(\frac{11}{4}-\frac{3}{2}\) = \(\frac{5}{4}\)
Now the expression has become, \(\left[ {3\frac{6}{7}\, ÷ \,\frac{{54}}{7}\, – \,\left\{ {3\, – \frac{5}{4}} \right\}} \right]+\frac{A}{4}=0\)
Solving the expression inside the curly braces {},
\(3-\frac{5}{4}=\frac{7}{4}\)
The simplified expression is, \(\left [3\frac{6}{7} \div \frac{54}{7}- \frac{7}{4}\right]+\frac{A}{4}=0\)
Solving expression of rectangular bracket,
\(3\frac{6}{7} \div \frac{54}{7}- \frac{7}{4}\) = \(\frac{27}{7} \div \frac{54}{7}- \frac{7}{4}\)
⇒ \(\frac{27}{7} \times \frac{7}{54}- \frac{7}{4}\) = \(\frac{1}{2}-\frac{7}{4}\)
⇒ \(\frac{-5}{4}\)
The expression is,
\(\frac{-5}{4}+\frac{A}{4}=0\)
⇒ A = 5
∴ The value of A is 5.
The average weight of 15 athletes and three trainers is 60 kg. When four more athletes of average weight 63 kg are included in the group, what is the average weight (in kg, rounded off to the nearest integer) of the group?
Given:- Average of 15 athletes and 3 trainers (18 people) is 60Kg.
To find:- Average when 4 more athlete included having average of 63
Formula used:- \(Average=\frac{Sum\ of\ observation}{No.\ of\ observation}\)
Calculation:
Average of 18 people (Athletes + trainer) = 60
By using formula of average, \(Average=\frac{Sum\ of\ observation}{No.\ of\ observation}\)
⇒Sum of their weights = 60 × 18 = 1080 kg
Average of 4 more athletes = 63
⇒sum of their weights = 63 × 4 = 252
Total weight of 19 athlete and 3 trainers = 1080+252= 1332 Kg.
New average = \(\frac{total\ sum}{total\ no.\ of\ people }\)
New average = \(\frac{1332}{22}\) = 60.54kg
New average = 60.54 kg
On rounding off, New average =61 kg
∴ The average weight of the group is 61 Kg.
If \(\frac{{{\rm{8r}}}}{{{{\rm{r}}^{\rm{2}}}\,{\rm{ – }}\,\,{\rm{8r}}\,{\rm{ + }}\,{\rm{1}}}}\, = \,\frac{1}{{14}}\), then the value of \(\left( {{\rm{r}}\,{\rm{ + }}\,\frac{1}{{\rm{r}}}} \right)\) is ______.
Given:-\(\frac{{{\rm{8r}}}}{{{{\rm{r}}^{\rm{2}}}\,{\rm{ – }}\,\,{\rm{8r}}\,{\rm{ + }}\,{\rm{1}}}}\, = \,\frac{1}{{14}}\)
To find:-
\(\left( {{\rm{r}}\,{\rm{ + }}\,\frac{1}{{\rm{r}}}} \right) \)
Calculation:
It is given that
\(\frac{{{\rm{8r}}}}{{{{\rm{r}}^{\rm{2}}}\,{\rm{ – }}\,\,{\rm{8r}}\,{\rm{ + }}\,{\rm{1}}}}\, = \,\frac{1}{{14}}\)
Dividing both the numerator and denominator by ‘r‘
⇒\(\frac{\frac{8r}{r}}{\frac{r^2-8r+1}{r}}=\frac{1}{14}\)
⇒\(\frac{8}{r-8+\frac{1}{r}}=\frac{1}{14}\)
On solving,
⇒14×8 = \(r-8+\frac{1}{r}\)
⇒\(\left(r+\frac{1}{r}\right) \) = 112+8
⇒ 120
∴ The value of \(\left(r+\frac{1}{r}\right) \) is 120.
After allowing 15% discount, a dealer wishes to sell a machine for ₹1,22,700. At what price must the machine be marked?
(Consider up to two decimals)
Given:
Discount percentage = 15%
Selling Price of machine = Rs 1,22,700
Concept Used:
Marked Price = Selling Price ÷ (100% – Discount%)
Calculation:
⇒ Putting all the values in MP formula,
⇒ Marked Price = 122700 ÷ (100% – 15%)
⇒ Marked Price = 122700 ÷ 85%
⇒ Marked Price = \(122700\times\frac{100}{85}\)
⇒ Marked Price = \(\frac{12270000}{85}=1,44,352.94\)
Hence, the price at which machine was marked is Rs 1,44,352.94
If \(\frac{{{x^2}}}{{{y^2}}}\, + \,\frac{{{y^2}}}{{{x^2}}}\, = \,7\), A possible of \(\frac{{{x^3}}}{{{y^3}}}\, + \,\frac{{{y^3}}}{{{x^3}}}\) is :
Given:
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}=7\)
Concept Used:
\((a+b)^2=a^2+b^2+2ab\)
\(a^3+b^3=(a+b)(a^2+b^2-ab)\)
Calculation:
⇒ \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=7\)
⇒ Adding both the sides \(2\frac{x}{y}\times\frac{y}{x}\),
⇒ \(\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\frac{x}{y}\times\frac{y}{x}=7+2\frac{x}{y}\times\frac{y}{x}\)
⇒ \((\frac{x}{y}+\frac{y}{x})^2=7+2\)
⇒ \(\frac{x}{y}+\frac{y}{x}=\sqrt9=3\)
⇒ Now using formula for a3 + b3,
⇒ \(\frac{{{x^3}}}{{{y^3}}}\, + \,\frac{{{y^3}}}{{{x^3}}}=(\frac{x}{y}+\frac{y}{x})(\frac{x^2}{y^2}+\frac{y^2}{x^2}-\frac{x}{y}\times\frac{y}{x})\)
⇒ Putting the values, we get
⇒ \(\frac{x^3}{y^3}+\frac{y^3}{x^3}=3\times(7-1)=18\)
∴ The required value of \(\frac{x^3}{y^3}+\frac{y^3}{x^3}\) is 18
The table below gives the numbers of three types of trees that were planted by government agencies in six consecutive years.
Year |
Banyan |
Neem |
Teak |
2013 |
30000 |
25000 |
15000 |
2014 |
35000 |
30000 |
5000 |
2015 |
35000 |
45000 |
10000 |
2016 |
40000 |
40000 |
25000 |
2017 |
45000 |
55000 |
35000 |
2018 |
55000 |
50000 |
40000 |
Given:
Teak trees planted in 2016 = 25000
Teak trees planted in 2017 = 35000
Concept Used:
Percentage Increase = \(\frac{Difference}{Trees}\times 100\)
Calculation:
⇒ Increase in teak trees in 2017 as compared to 2016 = Teak trees planted in 2017 – Teak trees planted in 2016
⇒ Increase in teak trees in 2017 as compared to 2016 = 35000 – 25000 = 10000
⇒ Percentage increase in teak trees from 2016 to 2017 = \(\frac{10000}{25000}\times100\)
⇒ \(\frac{10000}{25000}\times100=\frac{1000}{25}\)
⇒ Percentage increase = 40%
∴ The percentage increase in the number of teak tress that were planted in 2017 as compared to that by 2016 is 40%.
The speed of a boat is 12 km/h in still water. It goes 18 km downstream and comes back in a total time of 6 \(\frac{6}{7}\) hours. Find the speed of the stream (in km/h).
Given:
Speed of boat in still water = 12 km/h
Total distance for downstream = 18 km
Time taken to come back = \(6\frac{6}{7}\) hours
Concept Used:
\(Speed = \frac{Distance}{Time}\)
Downstream speed of the boat = Speed of boat + Speed of stream
Calculation:
Let the speed of the stream be x.
⇒ Downstream speed of the boat = 12 + x
⇒ Upstream speed of the boat = 12 – x
⇒ Time taken by the boat to cover 18 km downstream and 18 km upstream = \(6\frac{6}{7}\) = \(\frac{48}{7}\) hours
Now, using the formula for speed and putting the values, we get
⇒ \(\frac{18}{12 + x} + \frac{18}{12-x}=\frac{48}{7} \)
Taking LCM of denominators from LHS,
⇒ \(\frac{18\times(12-x)+18\times(12+x)}{(12+x)(12-x)}=\frac{48}{7} \)
By solving the equation,
⇒ \(432\times7=48\times(144-x^2) \)
⇒ \(48x^2=6912-3024\)
⇒ \(48x^2=3888\)
⇒ \(x^2=81\)
⇒ x = 9 km/h.
∴ The speed of the stream will be 9 km/h.
The pie-chart given below shows the monthly expenditures made by a family under different heads as percentages of the total monthly income of the family. If the total monthly income of the family is ₹70,000, then what is the amount that is left with the family after only the food and medical expenditures for the month are made?
Given:
Total monthly income of a family = Rs 70,000
Percentage of expenditure for food = 28%
Percentage of expenditure for house rent = 14%
Percentage of expenditure for child education = 25%
Percentage of expenditure for medical = 10%
Percentage of expenditure for entertainment = 8%
Percentage of expenditure for provident fund = 15%
Concept Used:
Income left = Total Income – Income Spend
Calculation:
Amount spent on food = 28% of 70,000
⇒ Amount spent on food = \(\frac{28}{100}\times70000\) = Rs 19,600
Amount spent on medical = 10% of 70,000
⇒ Amount spend on medical = \(\frac{10}{100}\times70000\) = Rs 7,000
⇒ Total amount spent in food and medical = Rs 19,600 + Rs 7,000 = Rs 26,600
⇒ Amount left after spending on food and medical = Rs 70,000 – Rs 26,600 = Rs 43,400
Hence, amount left with the family after only the food and medical expenditures for the month is Rs 43,400
Comment on “Notes 2”