Class : Class 8
Let us say that the total sum of weight is x
x/120 = 56
x = 6720
Now, let us say that the number girls is z
We know that the sum of the weight of the girls will be the sum of weight of the boys subtracted from the total sum.Let us say that the sum of weights of boys is y.
So now girls total weight sum will be : 6720-y/z = 50
6720-y = 50z
6720 = 50z + y
6720 - 50z = y(Equation 1)
Total sum of weight of boys will be
y/120 - z = 60
y = 7200 - 60z(Equation 2)
Subtracting both equations(1 and 2) :
y-y = 7200-60z - 6720z + 50z
0 = 480-10z
10z = 480
z = 48
Number of girls is 48.So, the number of boys will be 120-48 = 72
Post Your Answer
Post Your Answer
Post Your Answer
Post Your Answer
Post Your Answer
State ‘T’ for true and ‘F’ for false.
(i) Perimeter of equilateral triangle is four times its sides.
(ii) Without using the graph paper, we can find the area of a rectangle.
(iii) Distance covered along the boundary of a closed figure is called its area.
(i) | (ii) | (iii) | |
A | T | F | T |
B | F | T | F |
C | F | T | T |
D | F | F | T |
Post Your Answer
In the given figure, PQRS is a rhombus whose diagonal PR and QS are along coordinate axis and PR = 12 units and QS = 6 units. Now, if T is a point which is 5 spaces right and 2 spaces above S. Find :
(i) sum of abscissae of P and T.
(ii) sum of ordinates of Q, R and T.
(i) | (ii) | |
A | –1 | 2 |
B. | 1 | –2 |
C. | 1 | 2 |
D. | –1 | –2 |
explain pls
Ans 1:
Class : Class 9
Coordinates of the points-
P(-6, 0)
Q(0, -3)
R(6, 0)
S(0 ,3)
Now, T is 5 units to the right and 2 units above S, so its coordinates are T(0+5, 3+2)= T(5, 5)
i) Sum of abscissae (x-axis coordinate) of P and T = -6+5= -1
ii) Sum of the ordinates (y-axis coordinate) of Q, R and T= -3+0+5= 2
Hence, the correct answer is option A) (i) -> -1, (ii) -> 2