please solve it
very urgent
will mark u as brainiest
Answers
Answer:
3.
Volume of cuboid = l*b*h
If the length and breadth is doubled and height remains constant.
then , l = 2l , b = 2b
Then , new Volume of cuboid = (2l)*(2b)*h
=> Volume = 4 (l*b*h)
4.
Let the tea packets placed be n.
=> 50*30*0.2=n(10*6*4)
=> n = (50*30*0.2)/(10*6*4)
=> n = 5*5*0.5
=> n = 12.5
So nearly 12 tea packets can be placed
5.
Length (l) = 80 cm,
Breadth (b) = 30 cm,
Height (h) = 40 cm
/* According to the problem given,
Area of the paper required =
Area of the base + 2×Area of two side faces + Area of the back side
(Front side should be visible,So, it is not covered )
= lb+2×bh+lh
= 80×30+2×(30×40)+80×40
= 2400+2×1200+3200
= 2400+2400+3200
= 8000 cm²
Answer:
Q3. CASE 1
length = l
breath = b
height = h
volume1 = (lbh)
CASE 2
length = 2×l
breath = 2×b
height = h
volume 2 = 2l×2b×h
= 4(lbh)
=4volume
Thus, volume will become four timrs.
Q4. Measure of a packet = 10cm*6cm*4cm
Volume of a packet=(10∗6∗4)cm
3
=240cm
3
Measure of cardboard box = 50cm*30cm*0.2m
Volume of cardboard box = (50∗30∗20)cm
3
=30000cm
3
Number of packets =
Volumeofapacket
VVolumeofcardboardbo
=
240
30000
=125
So,there are 125 packets that can be placed in the given cardboard box.
Q4. l = 80 cm
b = 30 cm
h = 40 cm
Area of bottom = [ l×b = 80 ×30 = 2400 cm²]
Area of side face = b×h = 30 ×40 = 1200 cm²
Area of back face = l×b = 80×40 = 3200 cm²
= Needed area= area of base+area of back face+2×area of side face
=2400+3200+(2×1200)
=8000cm²
I hope it may help u...