Concept used:
Calculations:
\(\left[2\frac{1}{4}\div\frac{3}{4}\right]+\left[5\frac{2}{3}\div\frac{17}{13}\right]-\left[3\frac{1}{3}\times\frac{6}{5}\right]+[17.98\div6.2]+[1.25\div0.5]\)
⇒ \(\left[\frac{9}{4}×\frac{4}{3}\right] + \left[\frac{17}{3}×\frac{13}{17}\right]-\left[\frac{10}{3}×\frac{6}{5}\right]+2.9 +2.5\)
⇒ 3 + \(\frac{13}{3}\) – 4 + 5.4
⇒ 8.4 + 4.3 – 4
⇒ 8.7
∴ The Required simplified value is 8.7.
24Find the mean proportional between 144 and 225.
Formula used:
Mean proportion = √ab
Calculations:
= √ab
Mean proportion of 144 and 225
⇒ √(144 × 225)
⇒ √(32400)
⇒ 180
∴ Mean proportion of 144 and 225 is 180.
23Two numbers, A and B, are such that the sum of 10% of A and 8% of B is \(3\over5\)th of the sum of 12% of A and 16% of B. The ratio A ∶ B is:
Concept used:
a% of b = \(a \over 100 \) × b
Calculation:
10% of A + 8% of B = (\(\frac{3}{5}\)) (12% of A + 16% of B)
⇒ [(\(\frac{10A}{100}\)) + (\(\frac{8B}{100}\))] = (\(\frac{3}{5}\)) [(\(\frac{12A}{100}\)) + (\(\frac{16B}{100}\))]
⇒ (10A + 8B) = (\(\frac{3}{5}\)) (12A + 16B)
⇒ (10A + 8B) × 5 = 3 × (12A +16B)
⇒ 50A + 40B = 36A + 48B
⇒ 50A – 36A = 48B – 40B
⇒ 14A = 8B
⇒ A/B = 8/14
⇒ A : B = 8 : 14 = 4 : 7
∴ The ration of A : B is 4 : 7.
22A cone-shaped temple top has slant height of 11 m and radius of base is 10 m. Find the total cost for painting its curved surface area at the rate of Rs. 7/m2. [Use π = 22/7]
Given:
Slant height is 11 m,
Radius of base is 10 m,
Rate of painting is Rs. 7 per m2.
Formula used:
Total curved surface area = πrL
Calculations:
Total curved surface of Temple,
⇒ π.r.L
⇒ \(\frac{22}{7}\) × 11 × 10
⇒ 110π m2
Cost of painting,
⇒ 110π × 7 = 770 × \(\frac{22}{7}\)
⇒ Rs. 2420
∴ Cost of painting the curved surface area of Temple is Rs. 2,420.
21The diameter of a solid hemisphere is 10 cm. Find its total surface area (use π = 3.14).
Given:
Diameter of solid hemisphere is 10 cm.
Formula used:
Total surface area of hemisphere = 3πr2
Calculations:
Total surface area of hemisphere = 3πr2
Total surface area of hemisphere,
⇒ 3 × (\(\frac{22}{7}\)) × \((\frac{10}{2})\)2
⇒ 3 × (\(\frac{22}{7}\)) × \((\frac{100}{4})\)
⇒ 235.5 cm2
∴ Total surface area of hemisphere is 235.5 cm2.20Find the area of the given quadrilateral ABCD
Formula used:
Area of Triangle = \(\frac{1}{2}\) × Base × Height
Calculations:
According to the Question,
Area of quadrilateral ABCD,
⇒ Area of ∆ABC + Area of ∆ ADC
⇒ \(\frac{1}{2}\)× AC × BF + \(\frac{1}{2}\)× AC × DE
⇒ \(\frac{1}{2}\)× AC ( BF + DE )
⇒ \(\frac{1}{2}\) × 10 ( 5 + 3 )
⇒ 40 cm2
∴The area of quadrilateral ABCD is 40 cm2.
19The following bar graph shows the data of the number of students of different classes who like to eat different types of fruits
What percentage of the students of class 8 mentioned here like guavas?
Calculations:
According to question,In class 8 we have ,
No. of students who likes guava = 10
No. of students who likes Orange = 20
No. of students who likes Mangoes = 30
Total students in class 8 = 10 + 20 + 30 = 60
Now, percentage of no. of students who likes guava is = \(\frac{10}{60}\) × 100 = \(\frac{50}{3}\) = \(16\frac{2}{3}\)%
∴ Percentage of students who likes guavas is class 8 is \(16\frac{2}{3}\)%.
18Study the given table and answer the question that follows.
| Partners | Present share in percentage |
| Shourya | 18% |
| Tanmay | 22% |
| Bhanu | 25% |
| Dhanya | 35% |
If the company had Initally issued 10000 share between its four partners, and if Tanmay then sold 1000 of his shares to Dhanya, then how many shares did Dhanya finally have?
Given:
Shares among four members of a company is 18%, 22%, 25% and 35% to Shourya, tanmay, bhanu and dhanya respectively according to the table given.
The total shares initially is 10,000 and then tanmay sold 1000 shares to dhanya.
Calculations:
Dhanya share initially
⇒ 35% of 10,000
⇒\(\frac{35}{100}\) × 10,000
⇒ 3,500.
Now, total shares of dhanya after tanmay sold 1000 to her
⇒ 3,500 + 1,000
⇒ 4,500
∴ Total shares of Dhanya is 4500.
17(2x + 3y + 4z) (4x2 + 9y2 + 16z2 – 6xy – 12yz – 8xz) = ?
Formula used:
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Calculation:
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(2x)3 + (3y)3 + (4z)3 – 3 × 2x × 3y × 4z = (2x + 3y + 4z) (2x2 + 3y2 + 4z2 – 2x × 3y – 3y × 4z – 2x × 4z)
⇒ 8x3 + 27y3 + 64z3 – 72xyz = (2x + 3y + 4z) (4x2 + 9y2 + 16z2 – 6xy – 12yz – 8xz)
∴ The required value is 8x3 + 27y3 + 64z3 – 72xyz .
16A chord of 10 cm passes through the centre of a circle. What is the diameter of the circle?
Given:
A chord of 10 cm passes through the centre of a circle.
Concept used:
The chord passing through the centre of the circle is its Diameter.
Calculations:
The chord which passes through the centre of circle is it's diameter,
hence, The chord of length 10 cm is also it's diameter.
\(\therefore\) The Diameter of the circle is 10 cm.
15The average weight of a family of 4 members is 45 kg. The grandfather and the grandmother come to visit the family. Now the average weight of the 6 members of the family is 48 kg. Find the average weight of the grandparents (in kg).
Given:
The average weight of Family of 4 members is 45 kg.
After grandmother and grandfather's visit the average becomes 48 kg with total 6 members.
Concept Used:
Average = Sum of all numbers/Total number
Calculation:
There are four members in the family.
Total weight is
⇒ Average weight × total number of members
⇒ 45 × 4
⇒ 180 kg
Total weight after 6 members in the family
⇒ Average weight × total members
⇒ 48 × 6
⇒ 288 kg
Grandmother and Grandfather's age together is
⇒ 288 – 180
⇒ 108
Now, average age of grandmother and Grandfather
⇒ Sum of their ages / 2
⇒ \(\frac{108}{2}\) = 54 kg
∴ The Required average weight is 54 kg.
14What is the value of (4 sin3 x – 3 sin x + sin 3x)?
FORMULA USED :
\(( 4 sin^3x = 3 sin x- sin 3x)\)
CALCULATION:
\(( 4 sin^3x – 3 sinx + sin 3x)\) … (1)
Now,
We know that, \(4 sin^3x = 3 sinx -sin 3x\)
We have,
⇒ \(3 sinx – sin3x -3 sinx + sin3x\) = 0.
Hence, the value of the equation is 0.
13If 5 persons can prepare 5 tables in 5 days, then how long will it take 25 persons to prepare 25 tables?
GIVEN:
5 Persons prepare 5 Tables in 5 days.
FORMULA USED:
M1 × D1 × W2 = M2 × D2 × W1 Where,
M is no. of persons,
D is no. of days,
W is total work.
CALCULATION:
Here, M1 = 5 , D1 = 5 , W1= 5
And, M2 = 25, W2 = 25 . D2 = ?
Now, using formula, we get,
⇒ M1 × D1 × W2 = M2 × D2 × W1
⇒ 5 × 5 × 25 = 25 × D2 × 5
⇒ D2 = 5.
Hence, It will take 5 days to complete 25 tables.
12If m is a natural number, then the average of five consecutive natural numbers, starting with m, is:
FORMULA USED:
Average = sum of observation/ no. of observation.
CALCULATION:
Let the consecutive natural numbers starting with 'm' as,
⇒ m, ( m+1), ( m+2 ), ( m+3), ( m+4)
Sum of numbers,
⇒ m + (m+1) + (m+2) + (m+3) + (m+4) = (5m + 10)
⇒ Total numbers = 5.
⇒ Average = Sum/Total Numbers
⇒ Average = \(5 ( m+2)\over 5\) = (m+2).
Hence, The average is (m+2).
11The average weight of 18 boys in a class is 45 kg and that of the remaining 8 boys is 48.75 kg. Find the average weight (in kg) of all the boys in the class. (Correct to two decimal places)
GIVEN:
Average weight of 18 boys is 45 kg.
Average weight of 8 boys is 48.75 kg.
FORMULA USED:
Average = (Sum of Values)/(Number of Values)
CALCULATION:
Average of 18 boys = sum of 18 Boys/Number
⇒ 45 = sum of 18 boys/18
⇒ Sum of 18 boys = ( 45 × 18) = 1845
Average of remaining 8 Boys = Sum of 8 Boys/Number of boys.
⇒ 48.75 = Sum/8
⇒ Sum of the remaining 8 boys = 390.
Sum of all the boys = (sum of 18 boys + Sum of remaining 8 boys)
⇒ Sum of all the boys = (810 + 390) = 1,200.
⇒ Number of all the boys = (18 + 8) = 26.
Average = Sum of all the Boys/Number of Boys.
⇒ Average = 1200/26 = 46.15 kg.
Hence, Average Weight of all the boys is 46.15 kg.
10As part of Diwali offer, a jeweller allows a discount of 15%. Even after giving the discount, he makes a profit of 6.25%. Rani bought a gold chain which was marked at Rs. 5000 from this jeweller. Find the cost price of the chain for the jeweller.
GIVEN:
Discount is 15%
Profit is 6.25%
Marked Price is Rs. 5,000.
FORMULA USED:
Discount% = (Discount /M.P. )× 100
Discount = M.P. – S.P. where,
M.P. is Marked Price and S.P. is Selling Price.
Profit% = (Profit/C.P. ) × 100
Profit = S.P. – C.P.
C.P. = Cost Price.
CALCULATION:
Discount = 5000 – S.P.
⇒ Discount% = (Discount /M.P.) × 100
⇒ 15 = \(( 5000 – S.P.)\over5000\) × 100
⇒ 750 = 5000 – S.P.
⇒ S.P. = Rs. 4,250
Profit = (4,250 – C.P.)
⇒ Profit% = (Profit/C.P.) × 100
⇒ 6.25 = \(( 4250 -C.P.) \over C.P.\) × 100
⇒ 6.25 C.P. = 4,25,000 – 100C.P.
⇒ 106.25 C.P. = 4,25,000
⇒ C.P. = Rs 4,000.
Hence, Cost Price of the chain is Rs 4,000.
9If x > 0 and x4 + \(1\over{x^4}\) = 2207, what is the value of \(\rm \left(x^5+\frac{1}{x^5}\right)\)
GIVEN:
\(x^4 + 1/x^4 =2207\)
FORMULA USED:
\(( a + b) ^2 = a^2 + b^2 + 2ab.\)
where a and b = 1st and 2nd term respectively.
\(( a+b)^3 = a^3 +b^3 + 3.a.b( a+b)\) .
CALCULATION:
Now, \(( x^4 + 1/x^4)\) can be written as,
\((x^2 + 1/x^2)^2 = x^4 + 1/x^4 + 2.x^2.1/x^2\)
Here, a = x and b = 1/x respectively.
⇒ \(x^4 + 1/x^4 = (x^2 +1/x^2)^2 – 2\) ×\(x^2.1/x^2\)
⇒ 2207 = \(( x^2 + 1/x^2)^2\) – 2
⇒ \(( x^2 +1/x^2)^2\) = 2209.
⇒ \(( x^2 + 1/x^2)\) = 47 … (1)
Now, again \((x^2 +1/x^2)\) can be written as,
⇒ \(( x+1/x)^2 = x^2 +1/x^2 + 2.x .1/x.\)
⇒ \(( x^2 + 1/x^2)= ( x+1/x)^2\) -2
⇒ 47 = \(( x+1/x)^2\) -2
⇒ \(( x+1/x)^2\) = 49
⇒ (x +1/x) = 7 … (2)
Now, take the cube of equation (2) on both sides, we get
\((x +1/x)^3 = x^3 +1/x^3 +3.x.1/x (x+1/x)\)
Here a = x and b = 1/x.
⇒ 343 = \(( x^3 +1/x^3)\) + 3(7)
⇒ \((x^3 +1/x^3) \) = 322.
Now. \(( x^5 +1/x^5)\) can be written as,
⇒ \(( x^5 + 1/x^5 ) = ( x^2 +1/x^2) (x^3 +1/x^3) -( x+1/x)\)
⇒ ( 47 × 322 – 7) = 15127.
Hence, the value is 15,127.
8If the number 732XY is divisible by 70, then find the minimum value of \(\rm X+Y\over2\).
CONCEPT USED:
Divisibility rule :
10 = last digit should be zero.
7 =[Last 3 digit – remaining digits] should be multiple of 7.
CALCULATION:
Number = 732XY
Divisor = 10 × 7
According to the divisibility rule of 10,
Y should be zero.
⇒ Then the value of Y = 0.
And number = 732X0
Now, check the divisibility of 7,
⇒ 32X – 7
⇒ Here, possible values of X are = 1,2,3,4,5,6,7,8,9,0.
Now, put X =1,
321- 7 = 314.
⇒ 314/7 not divisible by 7.
Put X = 2,
322-7 = 315
⇒ 315/7 =45, which is divisible by 7.
Our required value of X = 2. Y = 0.
Now, the value of (X +Y) /2 = (2+0)/2 = 1.
Hence, the required value is 1.
7If x = 5t and y = \(1\over3\)(t +1), then the value of t for which x = 3y is:
GIVEN :
X = 5t
Y = 1/3 (t +1)
X = 3Y
CALCULATION :
We have, X = 3Y … (1)
So, put the value of X from equation (1) in X = 5t
⇒ 3Y = 5t … (2)
Now, put the value of Y i.e. Y = 1/3(t +1) in equation (2) we have,
⇒ 3 × 1/3 × ( t+1) = 5t
⇒ t +1 = 5t
⇒ 4t = 1
⇒ t = 1/4.
Hence, the value of t is 1/4.
6Two years ago, the population of a village was 1,000. The population increased by 10% is the first year and decreased by 10% in the second year, Find the present population.
GIVEN :
2 years ago,
The population of a village was 1,000.
1st year population increased by 10%
2nd year population decreased by 10%
FORMULA USED:
Present population = Actual population × (1±\(R \over 100\)), where
R is rate of increase and decrease and n is time.
CALCULATION:
Population after 1st year increase,
⇒ 1,000 × (1+10/100) = 1,000 × 11/10 = 1,100
Population after 2nd year decrease
⇒ 1,000 × 11/10 × (1-1/10) = 1,000 × 11/10 × 9/10 = 990.
Hence, Present population is 990.
5A cot was marked for Rs. 15,000 but sold for Rs. 13,500. What was the discount percentage offered on it?
Given:
Marked Price (M.P.) is Rs. 15,000
Selling Price (S.P.) is Rs. 13,500
Formula Used:
Discount % = \(discount\over M.P.\) × 100
Discount = M.P. – S.P.
Calculation:
Discount = ( M.P. – S.P.)
⇒ Discount = (15,000 – 13,500)
⇒ Discount = Rs 1,500
Now, Discount% = \(1,500\over 15,000\) × 100 = 10%.
Hence, Discount percentage is 10%.
4The table below gives the numbers of students from five different colleges who participated in the Olympiads of five different subjects during a given year.
| Subjects | College A | College B | College C | College D | College E |
| Hindi | 110 | 100 | 125 | 103 | 112 |
| English | 98 | 120 | 80 | 122 | 105 |
| Maths | 130 | 110 | 250 | 160 | 180 |
| Science | 100 | 100 | 150 | 200 | 80 |
| GK | 182 | 200 | 120 | 130 | 183 |
Find the average number of students per college who participated in the English Olympiad
GIVEN:
Five colleges A, B, C, D, and E
Five different subjects i.e. Hindi, English, Maths, Science, and G.K.
number of students participated in olympiad from 5 different colleges in 5 different subjects
FORMULA USED:
Average = \(sum of values\over number of values\) × 100
CALCULATION:
Number of students participated in English
⇒ From college A = 98
⇒ From college B = 120
⇒ From college C = 80
⇒ From college D = 122
⇒ From college E = 105
Hence, sum of all the students in English olympiad from 5 colleges,
⇒ 98 + 120 + 80 + 122 + 105 = 525
Number of colleges = 5
⇒ Average = 525/5 = 105.
Hence, the average no. of students per college participated is 105.
3In still water, the speed of a motorboat is 15 km/h. The boat travels 30 km downstream and returns in 4 hours and 30 minutes. The stream's speed (in km/h) is:
GIVEN:
Speed of a motorboat is 15 km/h
distance of boat in downstream is 30 km
distance of boat in upstream is 30 km
Total time is 4hrs 30 minutes
FORMULA USED:
Distance = Speed × Time
Relative speed in downstream = (speed of motorboat + speed of stream)
Relative speed in upstream = (speed of motorboat – speed of stream )
CALCULATION :
As 1hour = 60 minutes
To convert minutes into hours divide it by 60,
⇒ Time = (4 + 30/60) hours = 9/2 hours
Now, let the stream's speed be x
⇒ Downstream speed = ( 15 + x) km/h
⇒ Upstream speed = ( 15 -x ) km/h
Now, Distance = speed × time
⇒ Hence, \(Distance \over Speed \) = Time
⇒ \(30\over( 15 + x ) \) + \(30\over ( 15 -x )\) = 9/2
⇒ \( (15 +x +15 – x)\over225- x^2\) = 9/60 = \(30 \over( 225 – x^2)\)= \(9 \over 60\)
⇒ ( 200 = 225 – \(x^2\) ) = \(x^2\) = 25
⇒ x = 25.
Hence, the speed of the stream is 5 km/h.
2The marked price of a mobile phone is Rs. 18,000. It is sold with two successive discounts of 25% and 4%. An additional discount of 5% is offered for cash payment. The selling price of the mobile on cash payment is:
GIVEN:
Marked price is Rs. 18000
successive discounts is 25% and 4%
additional discount is 5%
FORMULA USED:
S.P. = (1 – d1/100) × (1 – d2/100) × M.P. where,
d1 and d2 are successive discount percentages respectively.
S.P. = Selling Price. and M.P. = marked price
CALCULATION:
Here, d1 = 25% and d2 = 4% and M.P. = 18,000
S.P. = (1-25/100) × (1-4/100) × 18000 = (1 -1/4) × (1 -1/25) × 18,000
⇒ S.P. = 3/4 × 24/25 × 18,000
⇒ S.P. = Rs. 12,960
Now,
There is an additional discount of 5% on cash payment,
Hence, additional discount of 5% in amount = 12960 × 5/100 = Rs. 648
S.P. after additional discount = (12,960 – 648) = Rs. 12,312.
Hence, Selling price of the mobile on cash payment is Rs. 12,312.
1By selling a fridge for Rs. 18200, Anu loses 15%. Find the cost price of the fridge. (correct to nearest whole number.)
GIVEN:
Selling price = Rs18200
loss% = 15
FORMULA USED:
loss % = \(Loss\over C.P.\) × 100 where, C.P. is Cost Price
Loss = C.P. – S.P. where C.P. is Cost Price and S.P. is Selling Price
CALCULATION:
Loss = C.P. – 18200
Loss % = \(( C.P. – 18200)\over C.P.\) × 100
⇒ 15 = \((C.P.. -18200)\over C.P.\) × 100
⇒ 15 C.P. = 100 C.P. – 1820000
⇒ 85C.P. = 1820000 ⇔ C.P. = Rs. 21,411.76 ≈ Rs. 21,412
Hence, Cost price of the fridge is Rs. 21,412.
0