In a school, a survey is taken on sports loving children as given in the following table. Any student who loves one sport is understood to dislike the other sport.
What is the difference between the number of boys and the number of girls in the school?
| Cricket | Soccer | |
| Boys | 542 | 438 |
| Girls | 358 | 372 |
Given:
Table showing the survey on students loving sports.
| Cricket | Soccer | |
| Boys | 542 | 438 |
| Girls | 358 | 372 |
Calculations:
Total number of boys = 542 + 438 = 980
Total number of girls = 358 + 372 = 730
Now, the difference between the number of boys and girls in the school = 980 – 730 = 250
\(\therefore\) The difference between no. of boys and girls in the school is 250.
24The number 2143251 is divisible by:
Given:
The number is 2143251
Concept used:
Divisibility law of 3 ⇒ A number is divisible by 3 if the sum of its digit is divisible by 3.
Divisibility law of 7 ⇒ A number is divisible by 7 if the difference between twice the unit digit of the given number and the remaining part of the given number should be a multiple of 7 or it should be equal to 0
Calculations:
Sum of digits of 2143251 = 2 + 1 + 4 + 3 + 2 + 5 + 1 = 18
Now, 18 is divisible by 3 therefore, the Number 2143251 is divisible by 3.
Here in the options, two numbers are not available so we don’t need to check divisibility from other numbers.
\(\therefore \)The number 2143251 is divisible by 3.
23If the ratio of the volumes of two cones is 11 ∶ 16 and the ratio of the radii of their bases is 3 ∶ 4, then the ratio of their heights will be:
Given:
The ratio between their radii of the bases = 3 : 4
The ratio between their volume = 11 : 16
Formula used:
The volume of the cone = \(\frac{1}{3}\)πr2h, where r is the radius of the base and h is the height of the cone.
Calculation:
Let V1 and V2 are the volumes of two cones
and H1 and H2 are their height.
V1 =\(\frac{1}{3}\) × π × (R1)2 × H1
V2 =\(\frac{1}{3}\) × π × (R2)2 × H2
According to the question,
V1 : V2 = 11 : 16
⇒ [\(\frac{1}{3}\) × π × (R1)2 × H1] : [\(\frac{1}{3}\) × π × (R2)2 × H2] = \(\frac{11}{16}\)
⇒ \(\frac{3²}{4²}\) × (H1 : H2)= \(\frac{11}{16}\)
⇒ \(\frac{9}{16}\) × (H1 : H2)= \(\frac{11}{16}\)
⇒ (H1 : H2) = 11 : 9
∴ The ratio between their heights is 11 : 9.
22In a class, the ratio of the number of boys to that of the girls is 11 ∶ 9. 30% of the boys and 20% of the girls passed an exam. Find the percentage of students who passed the exam.
Given:
The ratio of number of boys and girls = 11 : 9
30% of boys and 20% of girls passed the examination
Calculation:
Let the ratio of boys and girls be 11x and 9x.
Total number of students = 20x
According to the question,
Number of passed students = 11x × 30% + 9x × 20%
⇒ \(\frac{33x}{10}+\frac{18x}{10}\)
⇒ \(\frac{51x}{10}\)
Number of passed students = \(\frac{51x}{10}\)
Now, the percentage of passed students = \(\frac{\frac{51x}{10}}{20x}×100\)
⇒ 25.5 %
∴ The percentage of the passed student is 25.5 %.
21The mean of three numbers \(\frac{a}{3}\), b and c is the same as the mean of the numbers \(\frac{b}{3}\), \(\frac{c}{3}\) and x. The value of 3x is (where a, b, c are positive numbers):
Given:
The mean of three numbers \(\frac{a}{3}\), b and c is the same as the mean of the numbers \(\frac{b}{3}\), \(\frac{c}{3}\) and x.
Concept used:
Mean = \(\frac{Sum \space of \space observations}{No. \space of \space observation}\)
Calculations:
According to the question,
\(\frac{\frac{a}{3}+b+c}{3}=\frac{\frac{b}{3}+\frac{c}{3}+x}{3}\)
\(\Rightarrow \frac{a}{3}+b-\frac{b}{3}+c-\frac{c}{3}=x\)
\(\Rightarrow \frac{a+2b+2c}{3}=x\)
\(\Rightarrow 3x = a+2b+2c\)
\(\therefore\)The value of 3x = a + 2b + 2c.
20After successive discounts of 25% and 10% a shirt was sold for ₹480. What was the original price of the shirt (Nearest to a ₹)?
Given:
Selling Price (M.P) = Rs. 480
Two successive Discount = 25% and 10%
Formula used:
Net Discount = \((-X-Y+\frac{XY}{100})\)
S.P = M.P × \(\frac{(100-D\%)}{100}\)
Calculations:
Net Discount = \((-X-Y+\frac{XY}{100})\)
⇒ – 25 – 10 + 25 × \(\frac{10}{100}\)
⇒ – 35 + 2.5
⇒ – 32.5
Net discount% = 32.5 %
Selling Price(S.P) = Marked Price(M.P) × \(\frac{(100-D\%)}{100}\)
⇒ 480 = Marked Price × \(\frac{(100-32.5)}{100}\)
⇒ 480 = Marked Price × \(\frac{67.5}{100}\)
⇒ 480 = Marked Price × 0.675
⇒ Marked Price = \(\frac{480}{0.675}\)= 711.11 ≈ 711
∴ The original Price of the Shirt is Rs. 711.
19Find the sum of money which when increased by 15% becomes ₹19,320.
Given:
Final amount = Rs. 19,320
Gain% = 15 %
Calculations:
Let the amount be x.
Then, according to the question,
x + 15% of x = 19,320
\(\Rightarrow x + \frac{15x}{100}=19,320\)
\(\Rightarrow \frac{115x}{100}=19,320\)
\(\Rightarrow x =\frac{(19,320×100)}{115}\)
\(\Rightarrow\) x = 16,800
\(\therefore\) The initial sum of money is Rs. 16,800.
18The marked price of a toy was ₹4,875. Successive discounts of 28% and 30% were offered on it during a clearance sale. What was the selling price of the toy?
Given:
Marked Price (M.P) = Rs. 4, 875,
Two successive Discount = 28% and 30%
Formula used:
Net Discount = \((-X-Y+\frac{XY}{100})\)
S.P = M.P × \(\frac{(100-D\%)}{100}\)
Calculations:
Net Discount = \((-X-Y+\frac{XY}{100})\)
⇒ – 28 – 30 + 28 × \(\frac{30}{100}\)
⇒ – 58 + 8.4
⇒ – 49.6
Net discount% = 49.6 %
Selling Price(S.P) = Marked Price(M.P) × \(\frac{(100-D\%)}{100}\)
⇒ S.P = 4, 875 × \(\frac{(100-49.6)}{100}\)
⇒ S.P = 4, 875 × \(\frac{50.4}{100}\)
⇒ S.P = 4,875 × 0.504
⇒ S.P = Rs. 2, 457
∴ The selling Price of the Toys is Rs. 2,457.
17Harsh purchased a scooter for ₹48,000. He sold it at a loss of 15%. With that money he purchased another scooter and sold it at a profit of 22.5%. What is his overall loss/profit percentage?
Given:
C.P = Rs. 48,000
Loss% on first S.P = 15 %
Profit on second S.P = 22.5 %
Concept used:
S.P = \(\frac{(100-loss)}{100}×\) C.P
S.P = \(\frac{(100+profit)}{100}×\) C.P
Calculations:
First S.P = \(\frac{(100-15)}{100}×48,000\)
⇒ Rs. 40,800
Hence, First S.P = Second C.P = Rs. 40,800
Now, Second S.P = \(\frac{(100+22.5)}{100}×40,800\)
\(⇒\) Second S.P = Rs. 49,980
Now, Total profit = Second S.P – C.P = 49,980 – 48,000 = 1,980
Hence, total profit = Rs. 1, 980
Now, Profit % = \(\frac{1,980}{48,000}×100\)
\(⇒\) Profit % = 0.04125 × 100 = 4.125 %
\(\therefore\) Overall Profit % is 4.125%.
16Rohan scored 70 and 75 marks in the first two tests. How many marks should he score in the third test to get an average of 70 marks?
Given:
Rohan scored 70 and 75 in two exam .
Average given = 70
Concept used:
Average = Sum of all the terms/Total number of terms
Calculations:
Let the marks obtained in the third test be x.
According to the question,
\(\frac{70+75+x}{3}=70\)
\(\Rightarrow\) 145 + x = 210
\(\Rightarrow\) x = 65
\(\therefore \) Marks obtained in the third test are 65 marks.
15The table below provides information about the distribution of students from three different schools who are enrolled with three different coaching centres in a town.
| Coaching Centre | Total no. of students who have enrolled in the coaching centre | Percentage of students who have enrolled in the given coaching centre from different schools | ||
| L | M | N | ||
| P | 70 | 30 | 22 | 15 |
| Q | 50 | 32 | 24 | 29 |
| R | 300 | 25 | 34 | 18 |
35% of the students in Coaching centre R are female. 20% of the female students enrolled with centre R are school M. What is the number of male students from school M who have enrolled with Coaching centre R?
GIVEN:
The total number of students in coaching R = 300.
Percentage of students of school M enrolled in coaching R = 34
CALCULATION:
35% of the students in coaching R are female = 35% × 300 = 105 .
20% of the female enrolled with coaching R belongs to school M = 21
Total students are in the school M = 34% of 300 = 102.
Hence, female in school M = 21
No. of male students of school M who have enrolled with coaching R = (102 – 21) = 81.
Hence, the number of male students of school M who enrolled with coaching R = 81.
14Study the given pie-chart and answer the question that follows.
The pie-chart shows the data for A, B, O, AB blood groups of 240 donors.
 ÷ 2 = 160/3.</p>
<p>Hence, the ratio between O and the average of A and AB = 160/3 : 160/3 = 1:1.</p>
<p><strong>Hence, the ratio is 1:1.</strong></p>
_16572945737411.png”></p>
<p>The ratio of the donors of blood group ‘O’ to the average of the donors of groups ‘A’ and ‘AB’ is:</p>
<hr>62c84d5323daa1479e1fafc213<hr><span style=)
The value of tan 48°. tan 23°. tan 42°. tan 67° is
Given:
tan48° tan23° tan42° tan67°
Formula used:
tanθ × cotθ = 1
tan(90 – θ) = cotθ
Calculation:
⇒ The value can be rewritten as
tan48 tan42 tan23 tan67
⇒ tan48 tan(90 – 48) tan23 tan(90 – 23)
⇒ tan48 cot48 tan23 cot23
⇒ 1
Hence, the value is 1
12Find the area of the triangle whose sides are 5 cm, 12 cm and 13 cm.
GIVEN :
The sides of a triangle are 5 cm,12 cm and 13 cm
FORMULA USED:
Area of the triangle = \(\sqrt {s (s-a)(s-b) (s-c)}\) where a, b and c are the sides of the triangle respectively
and s is the semi- perimeter and s = (a + b + c) /2.
CALCULATION:
S = (5 + 12 + 13) /2 = 15 cm
Area of the triangle = \(\sqrt{15 (15 -5)(15-12)(15 -13)}\) = \(\sqrt{15 (10)(3)(2)}\) = \(\sqrt900\) = 30 cm2.
Hence, the area of the triangle is 30 cm2.
Alternate Method
\(13^2\) = 169 and \(( 5^2 + 12 ^2)\) = 169 .
Hence, as both are equal and the square of the bigger side is equal to the sum of the square of the smaller sides
\(13^2 = 5^2 + 12^2\) it is a right-angled triangle and hence, as in this \(h^2 = p^2 + b^2\) where h = hypotenuse, b= base and p= perpendicular.
⇒ here, h = 13, b = 12 and p = 5
⇒ Hence, Area of the triangle = 1/2 × b × h = 1/2 × 12 × 5 = 30 cm2 .
Hence, the area of the triangle is 30 cm2
11A man takes 16 days to complete a project. A woman does the same work in 12 days. They started the work together on alternate days and the man worked on the first day. How much time will they take to complete the work?
GIVEN :
Man completes the project in 16 days
Women complete the same work in 12 days
they worked on alternate days.
CALCULATION :
Man’s 1-day work = 1/16 days
Woman’s 1-day work = 1/12 days
(man + woman)’s work in 2 days = ( 1/16 + 1/12) = (4 + 3)/48 = \(\frac{7}{48}\)
⇒ In 6 pairs of 2 days they will complete = \(\frac{7}{48}\) × 6 = \(\frac{7}{8}\) i.e. after 6 × 2 = 12 days
Remaining work
(1 – 7/8) = 1/8 work is left .
On the 13th day, A will do the work,
⇒ 1/16 work done by A in 1 day
Then, remaining work = (1/8) – (1/16) = 1/16
On the 14th day, it is B’s turn,
⇒ 1/12 work done by B in 1 day
Hence, 1/16 work will be done in (12 × 1/16) = 3/4 day
⇒ Total days = (12 + 1 + 3/4) = \(13 \frac{3}{4} \) days or 55/4 days.
Hence, they will take 55/4 days to complete the work.
10The base of a 12 cm-high wooden solid cone has a circumference of 44 cm. Find its volume.
(use \(\pi \frac{{22}}{7}\))
GIVEN:
Height of wooden solid cone = 12cm
Circumference = 44cm
FORMULA USED:
Volume = \(\frac{1}{3}π r^2h\), where r is radius and h is height respectively
Circumference = \(2π r\)
CALCULATION :
Circumference = 2πr = 44
⇒ 2 × 22/7 × r = 44
⇒ r = 7 cm
The volume of the cone = \(\frac{1}{3}π r^2h\)
⇒ 1/3 × 22/7 × 7 × 7 × 12
⇒ 616 \(cm^3\)
Hence, the volume is 616 cm3
9The speed of a train is 108 km/h. The distance covered by the train in 11 seconds will be:
GIVEN :
Speed = 108 km/h
Time = 11 seconds
FORMULA USED:
Distance = Speed × Time
CALCULATION :
Speed = 108 km/h
To convert km/h into m/s multiply the number by 5/18.
speed = 108 × 5/18 m/s = 30 m/s.
Distance = Speed × Time = 30 × 11 = 330 m
Hence, the distance covered by the train is 330m.
8If x ∶ y = 1 ∶ 2, find the value of (2x + 4y) ∶ (x + 4y).
GIVEN:
x : y = 1 : 2
CALCULATION :
x/y = 1/2
Hence, on putting the value of x and y, we get
⇒ (2x + 4y) : (x + 4y) = (2 × 1 + 4 × 2) : ( 1 + 4 × 2)
⇒ (2x + 4 y) = 10 : 9.
Hence, the value is (2x + 4y) = 10 : 9.
7Triangles ABC and PQR are similar to each other. If m(\(\overline {BC} \)) = 5 cm, m(\(\overline {QR} \)) = 12.5 cm, and m(\(\overline {PQ} \)) = 6.25 cm, find m(\(\overline {AB}\)).
GIVEN:
Two Δ ABC and Δ PQR are similar to each other.
Side BC = 5 cm , Side QR = 12.5 cm, Side PQ = 6.25 cm
CONCEPT USED:
When two triangles are similar, their corresponding sides are in the same ratio.
CALCULATION :
Since Δ ABC is similar to Δ PQR
Hence, their corresponding sides are in the same ratio such as,
AB/PQ = BC/QR =AC/PR
⇒ Now, AB/PQ = BC/QR
⇒ AB/6.25 = 5/12.5
⇒ AB = 2.5 cm
Hence, the value of side AB is 2.5 cm.
6Compute the compound interest on ₹ 8,400 for one and half years at 12% rate of interest per annum, compounded half yearly.
GIVEN:
Sum = Rs 8400 \((1 +\frac{6}{100})^3\)
Time = One and a half years
Rate = 12% per annum compounded half early
FORMULA USED:
Compound Interest = P \(( 1 + \frac{r}{100})^n \) – P where n = time and P = principal or sum
When it is compounded half-yearly, then time becomes double i.e. = 3 years in this case
and the rate becomes half i.e. 6% in this case.
CALCULATION:
Principal = 8400
C.I. = 8400 \(( 1+ \frac{6}{100})^3\) – 8400
⇒ (8400 × 106/100 × 106/100 × 106/100) – 8400 = 1604.5
Hence, compound interest is Rs1604.5
5Which of the following options is divisible by 11?
GIVEN:
numbers divisible by 11
CONCEPT USED :
Divisibility of 11: Difference between the sums of the alternate digits of the given number is either 0 or divisible by 11.
CALCULATION :
Considering option (1) , we have
⇒ 28171 = (2 + 1 + 1) – (8 + 7) = (4 – 15) = -11, which is divisible by 11
⇒ Hence, 28171 is divisible by 11.
Considering option (2), we have
⇒ 27667 = (2 + 6 + 7) – (7 + 6) = (15 – 13) = 2, which is not divisible by 11
⇒ Hence, 27667 is not divisible by 11
Considering option (3), we have
⇒ 29817 = (2 + 8 + 7) – (9 + 1) = (17 – 10) = 7, which is not divisible by 11
Considering option (4), we have
⇒ 28196 = (2 +1 + 6) – (8 + 9) = (9 – 17) = -8 , which is also not divisible by 11
Hence, 28171 is divisible by 11.
4If the average marks of three batches of 30, 50 and 70 students is 55, 70 and 70 respectively, then the average marks of all the students is:
Given:
Average marks of 30, 50 and 70 students are 55, 70 and 70 respectively.
Formula used:
Average = Sum of all the values/Total Number of all the values.
Calculation:
Here, we have average marks of 30, 50 and 70 students are 55, 70 and 70 respectively
As, we know that Average = Sum of values / No. of values
Now,
Sum of marks of 30 students = 55 × 30 = 1650
Sum of marks of 50 students = 70 × 50 = 3500
Sum of marks of 70 students = 70 × 70 = 4900 .
Total sum of marks of all the three students = ( 1650 + 3500 + 4900 ) = 10,050
Total number of students = 30 + 50 + 70 = 150
Average = 10050 /150 = 67
Hence, the average mark of all the students is 67.
3If \((x – \frac{1}{x}) = \frac{7}{3}\), what is the value of \(\left( {{x^3} – \frac{1}{{{x^3}}}} \right)\)?
Given:
\((x – \frac{1}{x}) = \frac{7}{3}\)
Formula used:
\((a – b)^3 = a^3 – b^3 – 3ab(a -b)\) where a and b are 1st and 2nd term respectively.
Calculation:
\(( x-\frac{1}{x})^3= x^3 -\frac{1}{x^3} – 3x\frac{1}{x}(x- \frac{1}{x})\)
⇒ \((\frac{7}{3})^3\) = \( \left( {{x^3} – \frac{1}{{{x^3}}}} \right)\) – 3 × \(\frac{7}{3}\)
⇒ \(x ^3 – \frac{1}{x}^3\) = 532/27 = \(19 \frac{19}{27}\) .
⇒ Hence, the value is \(19\frac{19}{27}\).
2If a2 – 4a + 1 = 0, then the value of \({a^2} + a + \frac{1}{a} + \frac{1}{{{a^2}}}\) is:
GIVEN:
a2 – 4a + 1 = 0
FORMULA USED:
(a + b)2 = a2 + b2 + 2ab
Where a and b are 1st and 2nd term respective
CALCULATION:
a2 – 4a + 1 = 0
⇒ on taking a common we get,
⇒ a (a -4 +1/a ) = 0
⇒ a – 4 + 1/a = 0
⇒ a + 1/a = 4 ……………….( 1)
Now, on squaring both the sides we get,
( a + 1/a)2 = a2 + 1/a2 + 2 × a × 1
⇒ 16 = a2 + 1/a2 + 2
a2 + 1/a2 = 14 …………………….(2)
Now, on adding the equations (1) and (2) the value of a + 1/a + a2 + 1/a2 = ( 4 + 14) = 18.
Hence, the value is 18.
1The difference between a number and the square root of the number is 2. The number is:
GIVEN:
Difference between number and square root of the number = 2.
CALCULATION :
Let the number be x
then , According to the question
⇒ x – √x = 2 ————-(1)
On rearranging the equation we get
⇒ (x – 2) = √x
on squaring we get
⇒ (x – 2)2 = (√x)2
On apply the formula (a – b)2 = a2 + b2 – 2ab, we get
⇒ x2 + 4 – 2x (-2) = x
⇒ x2 – 5x + 4 = 0
⇒ x(x – 4) – (x – 4) = 0
⇒ (x – 1)(x – 4) = 0
⇒ x = 1 or x = 4
Now on putting the value of x in equation 1, we get
For x = 1,
⇒ x – √x = 2
⇒ 1 – √1 = 2
⇒ 1 – 1 = 2
⇒ 0 ≠ 2
For x = 4,
⇒ x – √x = 2
⇒ 4 – √4 = 2
⇒ 4 – 2 = 2
⇒ 2 = 2
So the required number is 4.
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