Given:
Increment in number = 10% of the number
Decrement in number = 20% of the new number
Concept used:
Net increase or decrease percent = {(Original number or Final number) – (Final number or original number) ÷ Original number} × 100
Calculation:
Let the original number be 100
When the number is increased by 10%
New Number = 100 + 10% of 100
⇒ 100 + 10
⇒ 110
Now, the number is decreased by 20%
Final number = New number – 20% of new number
⇒ 110 – 20% of 110
⇒ 110 – 22
⇒ 88
∴ Net Decrease in percentage = { (100 – 88) / 100} × 100
⇒ {12/100} × 100
⇒ 12%
∴ The net decrease percentage is 12%.
24In a class of 45 students, there are 20 girls who scored an average of 42 marks a test of 50 marks. If the class average is 36, what are the average marks of the boys?
Given:
Total students in class = 45
There are 20 girls who scored an average of 42 marks a test of 50 marks.
The class average is 36.
Calculations:
Let the average marks scored by boys be a.
According to the question
⇒ Average of class = 36 = (42 × 20 + 25 × a)/45
⇒1620 – 840 = 25a
⇒ a = 780/25 = 31.20
∴ Boys average marks are 31.20.
23If the circumradius of an equilateral triangle is 18 cm, then the measure of its in-radius is:
Given:
Circumradius of equilateral triangle = 18 cm.
Formula used:
Inradius of equilateral triangle = side/2√3
Circumradius of equilateral triangle = side/√3
Calculations:
Circumradius of equilateral triangle = side/√3
⇒ 18 = Side/√3
9 ÷ [5 + 7 ÷ {9 + 9 ÷ (9 + 9 ÷ 4)}] = ?
Given:
9 ÷ [5 + 7 ÷ {9 + 9 ÷ (9 + 9 ÷ 4)}] = ?
Concept Used:
Calculations:
9 ÷ [5 + 7 ÷ {9 + 9 ÷ (9 + 9 ÷ 4)}]
= 9 ÷ [5 + 7 ÷ {9 + 9 ÷ (9 + \(\frac{9}{4}\))}]
= 9 ÷ [5 + 7 ÷ {9 + 9 ÷ (\(\frac{45}{4}\))}]
= 9 ÷ [5 + 7 ÷ {9 + 9 × \(\frac{4}{45}\)}]
= 9 ÷ [ 5 + 7 ÷ { 9 + \(\frac{4}{5}\) }]
= 9 ÷ [ 5 + 7 × \(\frac{5}{49}\)]
= 9 ÷ [ 5 + \(\frac{5}{7}\) ]
= 9 × \(\frac{7}{40}\)
= \(\frac{63}{40}\)
\(\therefore\) The Required simplified value is \(\frac{63}{40}\).
21The value of \(x ^2+y^2,\) when x = 1, y = 2 is ________
Given:
x = 1, y = 2
Calculations:
x2 + y2
= (1)2 + (2)2
= 1 + 4
= 5
\(\therefore\)The value of x2 + y2 is 5.
20A man and a boy together can do a certain amount of digging in 24 days. Their speeds in digging are in the ratio of 4 : 3. How many days will the boy take to complete the work if engaged alone?
Given:
A man and a boy can together do a work in 24 days.
Efficiency can be written as Men: Boy = 4:3
Concept:
Number of Men × Number of days = constant
Calculations:
Efficiency can be written as Men: Boy = 4 : 3
So 1Man work = 4x (where x is any number)
1 Boy work = 3x
We know that,
Number of Men × Number of days = constant
(Man + Boy) × 24 = (Boy) × Days
\(\Rightarrow\) 7x × 24 = 3x × Days
Days = 56
\(\therefore\)Number of days boys will take to complete the work is 56.
19Study the given table and answer the question that follows.
The table shows the number of students doing various courses in various cities.
| City | MBA | MCA | M.Sc. | M.Com. | M.A. | Total |
| Ahmedabad | 1234 | 1384 | 1440 | 1289 | 1332 | 6679 |
| Bengaluru | 1156 | 1783 | 1874 | 1003 | 1340 | 7156 |
| Bhopal | 1187 | 1347 | 1532 | 1321 | 1486 | 6873 |
| Chennai | 1342 | 1473 | 1129 | 1765 | 1666 | 7375 |
| New Delhi | 1230 | 1098 | 1128 | 1865 | 1777 | 7098 |
| Hyderabad | 1456 | 1234 | 1556 | 1653 | 1789 | 7688 |
| Kolkata | 1239 | 1785 | 1865 | 1504 | 1762 | 8155 |
Which city has maximum number of student?
Given:
The no. Of students doing different courses in different cities.
Calculations:
According to question,
Kolkata has 8155 students as whole from all different courses.
\(\therefore\) Kolkata has maximum no. of students.
18The following Bar Chart shows the number of students enrolled in two Summer Camps A and B from 2014 to 2019. Study the charts carefully and answer the question that follows:
For the years 2014 to 2019, what is the ratio of the total number of students enrolled for Summer Camp A to that for Summer Camp B?
Given:
Bar Chart shows the number of students enrolled in two Summer Camps A and B from 2014 to 2019.
Calculations:
Sum of no. of students in camp A = 930
Sum of no. of students in camp B = 1410
Hence, Ratio between them is
= 930 : 1410
= 31 : 47
\(\therefore\)The ratio between students in camp A and camp B = 31 : 47.
17If x2 + 4y2 = 40, xy = 6 and x > 2y then the value of x – 2y is:
Given:
x2 + 4y2 = 40, xy = 6 and x > 2y .
Formula used:
a² + b² = (a – b)² + 2ab
Calculations:
According to question,
x2 + 4y2 = (x – 2y)2 + 4xy
\(\Rightarrow\) (x – 2y)2 = 40 – 4(6)
\(\Rightarrow\) (x – 2y) = (16)\(\frac{1}{2}\)
Hence, x – 2y = 4
\(\therefore\) The value of (x – 2y) is 4.
16The fourth proportional to 5, 8 and 30 is:
Concept Used:
a/b = c/d,
Where d is the fourth proportional of a, b, and c
Calculation:
Using the concept,
Let the fourth proportional be x
⇒ \(\frac{5}{8}\) = \(\frac{30}{x}\)
⇒ x = 6 × 8
⇒ x = 48
∴ Fourth proportional of 5, 8 and 30 is 48.
15Admissions into a commerce course of a university increase by 5% every year. If the number of admissions presently is 2400 into that course, after three years, the number of admissions is Nearest to an integer:
Given:
Principal amount = 2400
Rate = 5% per annum
Time = 3 years
Formula used:
Amount = P \((1+\frac{R}{100})\)Time
Calculations:
Amount = P \((1+\frac{R}{100})\)Time
= 2400\((1+\frac{5}{100})\)3
= 2400\((\frac{21}{20})\)3
= 2400 × \(\frac{9261}{8000}\)
= 2778.3 ≈ 2778
Hence, Number of admission after 3 year is nearest to 2778.
14A shopkeeper normally makes a profit of 20% in a certain transaction; he weights 900 g instead of 1 kg, due to an issue with the weighing machine. If he charges 10% less than what he normally charges, what is his actual profit or loss percentage?
Given:
A shopkeeper normally makes a profit of 20% in a certain transaction,
He weights 900 g instead of 1 kg, due to an issue with the weighing machine.
He charges 10% less than what he normally charges.
Formula used:
SP = \(\frac{100 – discount}{100}×CP\)
Calculations:
Let the cost price of 1 Kg of goods = Rs. 100
So, the selling price of 1 Kg of goods = 100 × 120/100 = Rs. 120
Cost price of 900 grams of goods = Rs. 90
According to question,
Shopkeeper charges 10% less what he normally charges
So, the new selling price = old selling price × (100 – 10)/100
⇒ New selling price = 120 × \(\frac{90}{100}\) =Rs. 108
So, profit = Rs. (108 – 90) = Rs. 18
So, profit % = (\(\frac{18}{90}\)) × 100 = 20%
Hence, Profit percentage is 20%.
13Calculate the area of the quadrilateral formed with the vertices (-3, 2), (5, 4),(7, -6) and (-5, -4).
Given:
Quadrilateral formed with the vertices (-3, 2), (5, 4),(7, -6) and (-5, -4).
Concept:
Area of quadrilateral ABCD = (1/2) ⋅ [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]
Calculations:
Let A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4) be the vertices of a quadrilateral ABCD.
Thus,
A(-3, 2) = (x1, y1)
B(5, 4) = (x2, y2)
C(7, -6) = (x3, y3)
D(-5, -4) = (x4, y4)
We know that,
Area of quadrilateral ABCD = (1/2) ⋅ [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]
Substituting the values,
= (\(\frac{1}{2}\)). {[-3(4) + 5(-6) + 7(-4) + (-5)2] – {[5(2) + 7(4) + (-5)(-6) + (-3)(-4)]}
= (\(\frac{1}{2}\)).[(-12 – 30 – 28 – 10) – (10 + 28 + 30 + 12)]
= (\(\frac{1}{2}\)) [-80 – 80]
= 160/2 {since area cannot be negative}
= 80
\(\therefore\)The area of the quadrilateral formed with the given vertices is 80 sq. units.
12The value of \({\sin4\theta+\sin2 \theta}\over{\cos 4\theta+\cos 2\theta}\) is:
Formula used:
sinA + sinB = 2 sin\(\frac{(A+B)}{2}\)cos \(\frac{(A-B)}{2}\)
cosA + cosB = 2 cos\(\frac{(A+B)}{2}\)cos\(\frac{(A-B)}{2}\)
Calculations:
\({\sin4θ+\sin2 θ}\over{\cos 4θ+\cos 2θ}\)
= \(\frac{2sin3θ cosθ}{2cos3θ cosθ}\)
= tan3θ
∴ The simplified value is tan3θ.
11Loan is to be repaid in two equal yearly instalments. If the rate of interest is 10% per annum, compounded annually, and each instalment is Rs.6,897, then find the total interest charged.
Given:
Amount of each installment = Rs. 6,897
Rate of interest compounded annually = 10%
Number of years (n) = 2
Concept used:
P = A/[1 + (R/100)]n
Interest = 2 × Installment – P
Calculation:
Let the loan amount be P.
P = 6897/[1 + (10/100)] + 6897/[1 + (10/100)]2
⇒ P = 6897/(1 + 1/10) + 6897/(1 + 1/10)2
⇒ P = 6897/(11/10) + 6897/(11/10)2
⇒ P = (6897 × 10)/11 + 6897/(121/100)
⇒ P = 68970/11 + 689700/121
⇒ P = 6270 + 5700
⇒ P = 11970
Interest = 2 × 6897 – 11970
⇒ 13794 – 11970 = 1824
∴ The total interest charged in this scheme is Rs. 1,824.
10Study the given graph and answer the question that follows.
The graph shows the number of appeared candidates and passed candidates ( in hundreds) in a test from six different institutions.
From which institution is the difference between the appeared candidates and passed candidates the maximum?
Given:
Calculations:
Difference between appeared students and passed students in institution A = 16 – 12 = 4
Difference between appeared students and passed students in institution B = 24 – 20 = 4
Difference between appeared students and passed students in institution C = 14 – 8 = 6
Difference between appeared students and passed students in institution D = 24 – 16 = 8
Difference between appeared students and passed students in institution E = 20 – 18 = 2
\(\therefore\)Difference between appeared students and passed students is maximum in institution D.
9
The average age of a man and his son is 56 years. The ratio of their ages is 11 : 5. What is the son's age?
Given:
The average of man and his son is 56 years.
The ratio of their ages is 11 : 5.
Calculations:
Total age = Average age × number of people
⇒ Total age of man and his son = 56 × 2 = 112 years
Ratio of their ages = 11 : 5
Let the ratio of age of man and son be 11x : 5x
⇒ 11x + 5x = 112 years or x = 7 years
∴ Age of son = 5x = 5 × 7 = 35 years.
8Air is pumped into a spherical balloon, its radius increases from 7 cm to 14 cm. Find the ratio of volume of the balloon's in both circumstances.
Given:
r : R = 7 : 14 = 1 : 2
Where r is radius before pumping and R is radius after pumping air.
Concept used:
In case of sphere,
v : V = r3 : R3
Calculations:
According to concept,
v : V = r3 : R3 = 73 : 143
⇒13 : 23 = 1 : 8
∴ The ratio of Volume of Sphere before and after pumping is 1:8.
7The average weight of 52 girl is 47 kg. What will be the sum of their weights?
Given:
The average weight of 52 girl is 47 kg.
Formula used:
Average = sum of observations / no. of observations
Calculations:
Let the sum of weight of 52 girls be x.
According to question,
Average = sum of observations / no. of observations
⇒ 47 = \(\frac{x}{52}\)
⇒ x = 2444
∴ The sum of weight of 52 girls is 2444 kg.
6A mobile is marked at a price 25% above its cost price. At what discount percentage should it be sold to make a 10% profit?
Given:
A mobile is marked at a price 25% above its cost price.
Profit = 10%
Formula used:
SP = \(\frac{100+Profit}{100}×CP\)
Calculations:
Let CP of the article be Rs. 100
MP of the article = Rs. 125
SP of the article
=
= 100 × (110/100)
= Rs. 110
Discount = 125 – 110 = 15
Discount percentage = 15/125 × 100 = 12%
Hence, the Discount percentage is 12%.
5Which of the following numbers is divisible by 33?
-
Concept used:
The divisibility rule of 3 states that a number is completely divisible by 3 if the sum of its digits is divisible by 3.
The divisibility rule of 11 states that the difference between the sum of the digits at even places and odd places is
either be 0 or divisible by 11.
Calculations:
Factors of the number 33 is
⇒ 33 = 3 × 11
So, the number which is divisible by 33 is also divisible by 3 and 11
Now, we go through option elimination method
Option (1) is 178750
⇒ 178750 = 1 + 7 + 8 + 7 + 5 + 0 = 28 [not divisible by 3]
Option (2) is 164461
⇒ 164461 = 1 + 6 + 4 + 4 + 6 + 1 = 22 [not divisible by 3]
Option (3) is 185592
⇒ 185592 = 1 + 8 + 5 + 5 + 9 + 2 = 30 [divisible by 3]
Option (4) is 171116
⇒171116 = 1 + 7 + 1 + 1 + 1 + 6 = 17 [not divisible by 3]
Now, Check 185592 is divisible by 11 or not
Take the sum of digits at odd places,
⇒ 1 + 5 + 9 = 15
Take the sum of digits at even places,
⇒ 8 + 5 + 2 = 15
Difference between the sum of the digits at even places and odd places is
⇒ 15 – 15 = 0
And, 0 is divisible by 11
So, 185592 is divisible by 33
∴ 185592 is divisible by 33
-
Suraj travels 124 km at a speed of 31 km/h by bus, 96 km at a speed of 16 km/h using a car, and another 105 km at 7 km/h by car. Find his average speed for the entire distance travelled.
Given:
Suraj travels 124 km at a speed of 31 km/h by bus,
96 km at a speed of 16 km/h using a car,
and another 105 km at 7 km/h by car.
Formula used:
Average speed = Total distance Travelled/Total time taken
Calculations:
Total distance travelled= 124 + 96 + 105
Total time taken = \(\frac{124}{31}+\frac{96}{16}+\frac{105}{7}\)
Average speed = \(\frac{124+96+105}{\frac{124}{31}+\frac{96}{16}+\frac{105}{7}}\)
Average speed = \(\frac{325}{4+6+15}\)
Average speed = 325/25 = 13 km/h
∴The average speed is 13 km/h.
3What is the surface area of a sphere whose diameter is 16.8 cm? [Use π = \(22\over7\)]
Given:
Diameter of sphere is 16.8 cm.
Formula used:
The surface area of a Sphere = 4πr²
Calculations:
Since the diameter is 16.8 cm, radius is 8.4 cm
Now,
The surface area of Sphere,
⇒ 4 × \(\frac{22}{7}\) × ( 8.4)²= 887.04 cm³.
∴ The Surface area of sphere is 887.04 cm³.
2Simplify the following expression.
\({(59\times59\times59)+(54\times54\times54)+(57\times57\times57)-3(59)(54)(57)} \over{(59+54+57)}\)
Given:
\({(59\times59\times59)+(54\times54\times54)+(57\times57\times57)-3(59)(54)(57)} \over{(59+54+57)}\)
Formula used:
a³ + b³ + c³ – 3abc = (a + b +c)(a² + b² + c² – ab – bc – ac)
Calculations:
\({(59\times59\times59)+(54\times54\times54)+(57\times57\times57)-3(59)(54)(57)} \over{(59+54+57)}\)
⇒ \(\frac{(54+59+57)(54²+59²+57²-(54×59 + 59×57 +57×54)}{(54+59+57)}\)
⇒ 54² + 59² + 57² – ( 3186 + 3363 + 3078)
⇒ 2916 + 3481 + 3249 – 9627
⇒ 9646 – 9627 = 19
∴ The simplified value is 19.
The marked price of a mobile phone is Rs. 36,000. A shopkeeper gives a discount of 11% on the marked price. Further, if a customer purchases it through a credit card then the final discount increases to 15%. Pooja purchases it with a credit card. How much does she pay?
Given:
Marked price of an article is Rs. 36, 000
Discount % = 11% and which further increases by 15%.
Concept used:
S.P = (100- Discount)/100
Calculation:
According to the Question,
Discount % = 11%
⇒ Final Discount % = 15%
Now, Selling price = \(\frac{(100-Discount\%)}{100}\)× 36, 000
⇒ S.P = \(\frac{(100-15)}{100}\)× 36, 000
⇒ S.P = \(\frac{85}{100}\)×36, 000
⇒ S.P = Rs. 30, 600
∴ Required S.P = Rs. 30,600.
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