Total Quantity of Lemon Juice
= 200 l - 200000 ml
Each student get = 250 ml
No. of students get the juice
= 200000 ml
250ml
= 800 ml
800 students got the juice

                   
The product of common factors of 40 and 56
= 2 \(\times\) 2 \(\times\) 2 = 8 and so, HCF (40, 56) = 8
Dividing by the common factor 2, (in 3 steps)
HCF = Product of common factors
 = 2 \(\times\) 2 \(\times\) 2 = 8

(i) An equilateral triangle has 3 equal sides.
Length of the wire = '12 s' cm
\(\therefore \) Length of each side =\(\frac { 12s }{ 3 } cm\)
Length of each side of the triangle =4s cm
(ii) A square has four equal sides.
\(\therefore \) Length of each side \(\frac { 12s }{ 3 } cm\)
Length of each side of the square = 3s cm.

Construction:
(i) Placed the centre of the protractor at the vertex '0', Lined up the ray \(\overline { OP } \) with the 0° line. Then draw and labelled a point Qat 90° mark on the
(a) inner scale (anticlockwise) and
(b) outer scale (clockwise)
(ii) Removed the protractor and draw \(\overline { OQ } \) to complete the angle.
Now < O = < POQ == < QOP= 90°.

Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJA and AJF \(\Rightarrow \)10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI => 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
\(\therefore \) Total 10 + 10 + 10 + 5 = 35 triangles.

(iii) Every part A, B, C, D, E, F, G, H, I, J, K, L, M, N, O and P are 16 triangles.
Combining two at a time. A & B, B & C, C & D, D & A, J & K, L & M, N & O, P & I are 8 triangles.
Taking the combination of 4, we have I, J, E & A; K, L, F and B, M, N, G and C, P, O, H and D are 4 triangles.
Taking 8 parts together we have QRS, RST, STQ, and TRQ are 4 triangles.
\(\therefore \) Total triangles = 16 + 8 + 4 + 4 = 32 triangles.

Pictograph of Number of computers sold

Here, 188 is nearer to 200 and 31 is nearer to 30.
The exact cost of 31 copies is 188 \(\times\) 31 = Rs. 5828 whereas,
The estimated cost of 31 copies = 200 \(\times\) 30 = Rs. 6000
Therefore, the estimated cost of 31 copies of Thirukkural books is Rs. 6000.

Correct answers given by Kaarmugilan and Kavitha are in the ratio 10:11
Total of 84 points will be divided into 10 + 11 = 21 equal parts
Points scored by Kavitha = \(\frac { 11 }{ 21 } \) \(\times\) 84 = 44 points
∴ Kavitha scored 44 points in the quiz.

40 in the ratio 6 : 14.
Dividing 40 into 6 + 14 = 20 equal parts and
\(40\times\frac{6}{20}=12\)
\(40\times\frac{14}{20}=28\)
\(\therefore\) The numbers are 12, 28.

True, because 72 is divisible by 72, and also 72 is divisible by 6.
Therefore this statement is true.

Let 3 odd natural numbers are 3, 7 and 9.
Sum = 3 + 7 + 9 = 19 is odd.
This statement is true.

562634 < 570000 < 572634

129 cm < 132 crn < 135 cm < 141 cm < 145 cm

Supplementary angle is 120°
Complementary angle is 30°

\(\angle ZYX \)

From the table m + 10=16 When m =6

Price of cool drinks per kg =P Rs.
Three times = 3P
5 Rs. more = 3P +5
\(\therefore \) Price of oil per kg = 3P +5

2n shows double the number of students who like icecream.

The pictograph for the mode of transport of students

Parallel and perpendicular lines

6 possibilities,
i) R - B - R - B  
ii) R - R - B - B 
iii) B - R - R - B 
iv) B - R - B - R  
v) B - B - R - R  
vi) R - B - B - R
(R- RED, B- BLUE)

(i) Obtuse angle
(ii) Zero angle
(iii) Straight angle
(iv) Acute angle
(v) Right angle

2s - 6 = 30
2s = 30 + 6
2s = 36
s = 36/2
s = 18

Place value chart is given by

\(\Rightarrow\) The number with more digits is the greater number
Step 1: \(\therefore\) 128435 is the larger number and 6348 is the least number
Step 2:  For the remaining 5 digit numbers we can compare the left most digits and find 25840 > 21354 > 10835
\(\therefore\)  The descending order:
128435> 25840 > 21354 > 10835> 6348

3:2
\(\frac{3}{2}\times\frac{2}{2}=\frac{6}{4}=6:4\)
\(\frac{3}{2}\times\frac{3}{3}=\frac{9}{6}=9:6\)
6 : 4 and 9 : 6 are equivalent to 3 : 2

(i) Vertical Bar Graph.

(ii) Horizontal Bar Graph.

Let karnan's age be x years
Gopals age is x-8 years
Given sum of their ages is 30.
i.e. x+(x-8) = 30
Now we will form the table to find thier sums =30

For x =19(Karnan's age) the sum = 30
\(\therefore \) Karnan's age = 19 years

(i) Indian System :Two crore fifty nine lakh forty one thousand nine hundred
International System : Twenty five million nine hundred forty one thousand nine hundred
(ii) 5,50,500
(iii) Eighty six crore forty lakh seven hundred thirty
(iv) Ninteen million eight hundred eighty eight thousand eight hundred
(v) Indian System : 60,53,100 - Sixty lakh fifty three thousand one hundred
International System : 6,053,100 - Six million fifty three thousand one hundred

Step 1: Complete the corners with smaller numbers 1, 2 and 3.

Step 2: The side having smallest numbers 1 & 2 are to be filled with the greatest number 6, the second smallest 1 & 3 side to be filled with the second largest 5 at the middle and so on.

The magic sum is 1 + 6 + 2 = 2 + 4 + 3 = 3 + 5 + 1 = 9. Some other ways are given below

The magic sum = 1 + 6 + 3 = 3 + 2 + 5 = 5 + 4 + 1 = 10

The magic sum 6 + 1 + 4 = 4 + 5 + 2= 2 T 3 + 6 = 11.

The magic sum 4 + 3 + 5 = 5 + 1 + 6= 6 + 2 + 4 = 12.

Ratio of red beads: blue beads = 35: 30
Fraction of 35 : 30  \(=\frac{35}{30}\)
= 7:6
Also given that the bracelet made is smaller in size
∴ The possible ways are red beads: blue beads
The equivalent ways are
\( \frac{7 \times 2}{6 \times 2}=\frac{14}{12}=14: 12 \)
\( \frac{7 \times 3}{6 \times 3}=\frac{21}{18}=21: 18 \)
\( \frac{7 \times 4}{6 \times 4}=\frac{28}{24}=28: 24\)
= 7 : 6 (or) 14: 12 (or) 21 : 18 (or) 28 : 24
In 4 different way can she made.