25. volume & surface areas c&add=t [www.onlinebcs.com]

VOLUME AND SURFACE AREAS 【819】
BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE
I. AvqZvKvi Nbe¯‘ :
ˆ`N©¨ = l, cÖ¯’ = b Ges D”PZv = h GKK n‡j
 AvqZb = lbh Nb GKK
 c„ôZ‡ji †ÿÎdj = 2(lb + bh + lh) eM© GKK
 KY© = l2
+ b2
+ h2
GKK
l 
h
b
II. NbK :
Nb‡Ki cÖwZwU avi a GKK n‡j,
 AvqZb = a3
Nb GKK
 c„ôZ‡ji †ÿÎdj = 6a2
eM© GKK
 KY© = 3a GKK
a
a
a
III. wmwjÐvi :
f‚wgi e¨vmva© r Ges D”PZv h n‡j,
 AvqZb = r2
h Nb GKK
 eµZ‡ji †ÿÎdj = 2rh eM© GKK
 mgMÖZ‡ji †ÿÎdj = (2rh + 2r2
) eM© GKK
= 2r (r + h) eM© GKK
r
h
r
IV. †KvYK :
f‚wgi e¨vmva© r Ges D”PZv h n‡j,
 †njv‡bv D”PZv l = h2
+ r2
GKK
 AvqZb =
1
3
r2
h Nb GKK
 eµZ‡ji †ÿÎdj = rl eM© GKK
 mgMÖZ‡ji †ÿÎdj = (rl + r2
) eM© GKK
h l
r
V. †KvY‡Ki UzKivi †ÿ‡Î :
hLb †Kvb †KvY‡K f‚wgi mgvšÍiv‡j †K‡U †bIqv nq ZLb f‚wg †_‡K †K‡U †bIqv Zj ch©šÍ Ask‡K
†KvY‡Ki UzKiv e‡j|
f‚wgi e¨vmva© R, KvUv Z‡ji e¨vmva© r Ges D”PZv h n‡j,
 AvqZb =
h
3
(R2
+ r2
+ Rr) Nb GKK
 †njv‡bv D”PZv l = (R – r)2
GKK
 eµZ‡ji †ÿÎdj = l (R + r) eM© GKK
 mgMÖZ‡ji †ÿÎdj =  [R2
+ r2
+ l (R + r)] eM© GKK
r
h
R
l
VI. †MvjK :
†Mvj‡Ki e¨vmva© r n‡j,
 AvqZb =
4
3
r2
Nb GKK
 Z‡ji †ÿÎdj = 4r2
Nb GKK
VII. Aa©‡MvjK :
Aa©‡Mvj‡Ki e¨vmva© r n‡j,
 AvqZb =
2
3
r3
Nb GKK
 eµZ‡ji †ÿÎdj = 2r2
eM© GKK
 mgMÖZ‡ji †ÿÎdj = 3r2
eM© GKK
r
VIII. wcivwgW :
 AvqZb =
1
3
 f‚wgi †ÿÎdj  D”PZv
 mgMÖZ‡ji †ÿÎdj = f‚wgi e¨vmva© + cvk¦©Zj¸‡jvi †ÿÎdj
GB Aa¨v‡qi ¸iæZ¡c~Y© Z_¨ I m~Î
25 Volume and Surface Areas

【820】 BANK MATH BIBLE
UvBc bs UvBc Gi bvg cÖkœ b¤î
1
†ÿÎdj ev AvqZb †_‡K ˆ`N©¨, cÖ¯’,
D”PZv wbY©q
5, 7, 11, 14, 17, 29, 49, 54, 56, 69, 70, 88, 91, 106, 107, 124, 142, 143, 144, 148,
176, 207, 221, 222, 236, 237, 242, 243, 248, 249, 250, 258, 286, 294, 304
2 †ÿÎdj, AvqZb wbY©q
3, 4, 8, 13, 15, 16, 38, 40, 43, 44, 45, 46, 52, 53, 58, 59, 60, 62, 63, 64, 65, 67,
68, 74, 75, 79, 83, 85, 103, 105, 108, 121, 149, 152, 164, 165, 166, 173, 256,
279, 285, 303
3 LiP msµvšÍ mgm¨v 10, 18, 19, 48, 51, 57, 61, 112, 115, 287
4 wbw`©ó AvqZ‡bi g‡a¨ ivLv e¯‘i msL¨v wbY©q 6, 22, 23, 25, 39, 78, 81, 86, 87, 90, 119, 120, 141, 194, 223, 228, 240, 251, 263
5 gvwU, cvwbi D”PZv e„w× wbY©q 9, 27, 28, 30, 31, 76, 102, 138, 139, 146, 147, 244
6 Rj c~Y© ev Lvwj Ki‡Z mgq wbY©q 33, 34, 35, 36, 135, 136, 148, 196
7 †ÿÎdj, AvqZb e„w× msµvšÍ 12, 26, 41, 42, 55, 56, 84, 93, 101, 116, 127, 131, 132, 179, 180, 189, 211, 215,
218, 227, 300, 301
8 AbycvZ wbY©q
37, 89, 92, 94, 95, 96, 97, 98, 99, 109, 113, 117, 118, 122, 123, 125, 126, 128,
129, 130, 160, 178, 181, 183, 184, 185, 188, 195, 202, 213, 214, 216, 219, 220,
231, 232, 238, 245, 255, 260, 265, 267, 270, 290, 291, 296, 297
9 IRb wbY©q 32, 47, 50, 77, 155, 203
cÖkœ b¤î:
1. A cuboid has — edges. (GKwU Nb‡ÿ‡Îi †gvU av‡ii msL¨vÑ)
a 4 b 8 c 12 d 16 c
 mgvavb : †Kvb Nbe¯‘i avi n‡jv
Gi †h‡Kvb `ywU Z‡ji †Q`‡b Drcbœ
†iLv ev Nbe¯‘i evû|
D
E
H G
A B
C
F
†gvU avi (AB, BC, CD, DA, EF, FG, GH, HE, AH, BG,
CF, DE) = 12wU|
2. 1 litre is equal to (1 wjUvi = ?)
a 1 cu. cm b 10 cu. cm
c 100 cu. cm d 1000 cu. cm d
3. *A rectangular water tank is 8 m high, 6 m long and 2.5
m wide. How many litres of water can it hold? (GKwU
AvqZvKvi cvwbi U¨vs‡Ki D”PZv 8 m, ˆ`N©¨ 6 m I cÖ¯’ 2.5 m|
GwU KZ wjUvi cvwb aviY Ki‡Z cv‡i?) [www.examveda.com]
a 120 litres b 1200 litres
c 12000 litres d 120000 times d
 mgvavb : AvqZvKvi U¨vs‡Ki AvqZb = ˆ`N¨  cÖ¯’  D”PZv
= 6 m  2.5 m  8 m = 120 m3
= 120  1000 liters [ 1 m3
= 1000 litres]
= 120000 litres
4. *The dimensions of a cuboid are 7 cm, 11 cm and 13 cm.
The total surface area is (GKwU Nb‡ÿ‡Îi avi¸‡jv 7 cm, 11
cm I 13 cm. n‡j mgMÖ c„‡ôi †ÿÎdjÑ) [www.examveda.com]
a 311 cm2
b 622 cm2
c 1001 cm2
d 2002 cm2
b
 mgvavb : ˆ`N©¨, l = 13 cm; cÖ¯’, b = 11 cm; D”PZv, h = 7 cm
mgMÖ c„‡ôi †ÿÎdj = 2(lb + bh + lh)
= 2  (13  11 + 11  7 + 13  7) cm2
= 622 cm2
5. *A closed aquarium of dimensions 30 cm  25 cm  20
cm is made up entirely of glass plates held together
with tapes. The total length of tape required to hold the
plates together (ignore the overlapping tapes) is (Kuv‡Pi
KZK¸‡jv †cøU †U‡ci mvnv‡h¨ jvwM‡q 30 cm  25 cm  20 cm
mvB‡Ri GKwU e× A¨vKzqvwiqvg ˆZwi Kiv n‡jv, †cøU¸‡jv‡K
ci¯ú‡ii mv‡_ AvU‡K ivL‡Z KZ ˆ`‡N©ï †Uc jvM‡e?)
[www.examveda.com]
a 75 cm b 120 cm
c 150 cm d 300 cm d
 mgvavb : AB = CD = EF = GH = 30 cm
BG = CF = AH = DE = 25 cm
AD = BC = EH = FG = 20 cm
H
A
D C
E F
G
B
eëüZ Kuv‡P †cøU¸‡jv n‡jv :
ABCD = EFGH
DCFE = ABGH
BCFG = ADEH
cvkvcvwk `ywU †cøU‡K GK‡Î †Uc w`‡q †Rvov w`‡q jvMv‡bv nq|
†hgb : ABCD I CDEF †K CD †iLv eivei †Uc jvwM‡q †Rvov
†`qv nq| ZvB Nb‡ÿÎiƒc A¨vKzqvwiqv‡gi cÖ‡Z¨K evû eivei †Uc
jvMv‡bv n‡q‡Q|
†U‡ci †gvU ˆ`N©¨ = 4  (AB + BG + BC)
= 4  (30 + 25 + 20) cm = 300 cm
Dr. R.S. AGGARWAL m¨v‡ii eB‡qi c~Y©v½ evsjv mgvavb
wiwfkb e· cieZx©‡Z †h cÖkœ¸‡jv Avcbvi wiwfkb Kiv cÖ‡qvRbÑ †m¸‡jvi b¤î wj‡L ivLyb 
GKB wbq‡gi AsK¸‡jv GK mv‡_ Abykxjb Ki‡Z

6. The dimensions of a room are 15 m, 10 m and 8 m. The
volume of a bag is 2.25 m3
. The maximum number of
bags can be a accommodated in the room is. (GKwU K‡ÿi
avi¸‡jv 15 m, 10 m I 8 m| GKwU e¨v‡Mi AvqZb 2.25 m3
| H
K‡ÿ GB gv‡ci m‡e©v”P KZ¸‡jv e¨vM ivLv hv‡e?)
a 531 b 533
c 535 d 550 b
 mgvavb : K‡ÿi AvqZb = 15  10  8 m3
= 1200 m3
 e¨v‡Mi msL¨v =
1200
2.25
= 533.33 = 533wU
7. A rectangular water reservoir contains 42000 litres of
water. If the length of reservoir is 6 m and breadth of
the reservoir is 3.5m, then the depth of the reservoir
will be (AvqZvKvi GKwU Rjvav‡ii 4200 wjUvi cvwb a‡i| hw`
Rjvav‡ii ˆ`N©¨ I cÖ¯’ h_vµ‡g 6 m I 3.5 m nq, Z‡e MfxiZvÑ)
a 2 m b 5 m c 6 m d 8 m a
 mgvavb : Rjvav‡ii AvqZb = 42000 litres =
4200
1000
m3
= 42 m3
MfxiZv h n‡j AvqZb = 6  3.5  h
cÖkœg‡Z, 6  3.5  h = 42 = h =
42
6  3.5
=
7
3.5
= 2 m
8. *A cistern 6 m long and 4 m wide contains water up to a
depth of 1 m 25 cm. The total area of the wet surface is.
(6 m `xN© I 4 m cÖk¯Í GKwU †PŠev”Pvq cvwbi MfxiZv 1 m 24
cm| †PŠev”Pvi Af¨šÍ‡i †fRv ev Av`ª© Z‡ji †ÿÎdj KZ?)
[www.examveda.com; www.indiabix.com]
a 49 m2
b 50 m2
c 53.5 m2
d 55 m2
a
 mgvavb : 1 m 25 cm = 1 m +
25
100
m
= 1.25 m
ABCDEFGH As‡k cvwb Av‡Q|
A
B
E
F G
H
1.5 cmD
6m
4m
AB = CD = FG = EH = l = 6 m
BC = DA = EF = GH = 6 = 4 m
AF = BG = CH = DE = h = 1.25 cm
EFGH Ask Db¥y³ hvi †ÿÎdj = lb
 cvwb Øviv †fRv Z‡ji †ÿÎdj = 2(lb + bh + lh) – lb
= lb + 2 (bh + lh)
= lb + 2h (l + b)
= [6  4 + 2  1.25  (6 + 4)]m2
= (24 + 25) m2
= 49 m2
9. *A boat having a length 3 m and breadth 2 m is floating
on a lake. The boat sinks by 1 cm when a man gets on
it. The mass of man is (3 m `xN© I 2 m PIov GKwU †bŠKv
†Kvb Lv‡ji cvwbi DcwiZ‡j fvmgvb| GKRb e¨w³ J †bŠKvq
IUvi d‡j †bŠKvwU 1 cm Mfx‡i †b‡g hvq| H e¨w³i fi KZ?)
a 12 kg b 60 kg
c 72 kg d 96 kg b
 mgvavb : †bŠKvi Wz‡e hvIqv As‡ki
AvqZb = AcmvwiZ cvwbi AvqZb
= l  b  h
b = 2m
h = 1cm
= 0.01m
l = 3m
= 3  2  0.01 m3
= 0.06 m3
 AcmvwiZ cvwbi fi = AcmvwiZ cvwbi AvqZb  NbZ¡
= 0.06  1000 kg = 60 kg
AvwK©wgwW‡mi bxwZ Abyhvqx, fvmgvb Ae¯’vq e¨w³i fi = AcmvwiZ
cvwbi fi = 60 kg
10. *A water tank is 30 m long, 20 m wide and 12 m deep. It
is made of iron sheet which is 3 m wide. The tank is
open at the top. If the cost of the iron sheet is 10 Tk.
per metre, then the total cost of the iron sheet required
to build the tank is. (30 m `xN©, 20 m PIov I 12 m Mfxi
GKwU cvwbi U¨vsK 3 m PIov †jvnvi cvZ Øviv ˆZwi| U¨vs‡Ki
DcwifvM Db¥y³| cÖwZ wgUv‡i †jvnvi LiP 10 UvKv n‡j U¨vsKwU
wbg©v‡Y †jvnvi cv‡Zi Dci †gvU LiP KZ?) [www.examveda.com]
a 6000 Tk. b 8000 Tk. c 9000 Tk. d 10000 Tk. a
 mgvavb : Dc‡ii Ask (lb) Db¥y³, ZvB U¨vs‡Ki †ÿÎdj
= lb + 2h (l + b) = [30  20 + 2  12  (30 + 20)]m2
= 1800 m2
 eëüZ cv‡Zi †ÿÎdj = 1800 m2
eëüZ cv‡Zi ˆ`N©¨ =
1800
3
m = 600 m
†gvU LiP = ˆ`N©¨  GKK ˆ`‡N©¨ LiP
= 600  10 UvKv = 6000 UvKv
11. Given that 1 cu. cm of marbel weighs 25 gms, the
weight of a marble block 28 cm in width and 5 cm thick
is 112 kg. The length of the block is (†Ìqv Av‡Q, 1 cm3
gv‡e©‡ji IRb 25 gm| 28 cm PIov I 5 cm cyiæ GKwU gv‡e©j
eø‡Ki IRb 112 kg n‡j Gi ˆ`N©¨ KZ?) [www.examveda.com]
a 26.5 cm b 32 cm c 36 cm d 37.5 cm b
 mgvavb : eø‡Ki IRb = 112 kg = 112  1000 gm
= 112000 gm
eø‡Ki ˆ`N©¨ l cm n‡j AvqZb = l  28  5 cm3
= 140 l cm3
 1 cm3
AvqZ‡bi eø‡Ki IRb =
112000
140 l
gm =
800
l
gm
cÖkœg‡Z,
800
l
= 25  l =
800
25
 l = 32 cm
12. Half cubic metre of gold sheet is extended by
hammering so as to cover an area of 1 hectare. The
thickness of the sheet is (
1
2
m3
AvqZ‡bi GKwU †mvbvi cvZ‡K
nvZzwo w`‡q wcwU‡q 1 †n±i we¯Í…Z Kiv n‡jv| we¯Í…Z Ae¯’vq cv‡Zi
cyiæZ¡ KZ?)
a 0.0005 cm b 0.005 cm c 0.05 cm d 0.5 cm b
 mgvavb : 1 †n±i = 10000 m2
†cUv‡bvi d‡j AvqZ‡bi †Kvb cwieZ©b nq bv|
we¯Í…Z Ae¯’vq cyiæZ¡ w n‡j, AvqZb = c„ôZj  cyiæZ¡
= 10000 w m3
cÖkœg‡Z, 10000 w =
1
2
 w =
1
2  10000
m =
0.5
10000
m
= 0.00005 m = 0.00005  100 cm = 0.005 cm
13. *In a shower, 5 cm of rain falls. The volume of water
that falls on 1.5 hectares of ground is : (†Kvb kvIqv‡i 5
cm cyiæ e„wócvZ nq| 1.5 †n±i we¯Í…Z Rwg‡Z cwZZ e„wói cvwbi
AvqZb KZ?) [www.examveda.com; www.indiabix.com]
a 75 cu. m b 750 cu. m
c 7500 cu. m d 7500 cu. m b
 mgvavb : AvqZb = Rwgi †ÿÎdj  e„wói cyiæZ¡
= 1.5 †n±i  5 cm
= (1.5  10000 m2
) 



5
100
m = 750 m3

14. The breadth of a room is twice its height and half its
length. The volume of the room is 512 cu. m. The
length of the room is (GKwU K‡ÿi cÖ¯’ D”PZvi wØ¸Y I ˆ`‡N©ï
A‡a©K| KÿwUi AvqZb 512 m3
n‡j ˆ`N©¨Ñ) [www.examveda.com]
a 16 m b 18 m c 20 m d 32 m a
 mgvavb : awi, ˆ`N©¨ = x;  cÖ¯’ =
x
2
cÖ¯’ = 2  D”PZv  D”PZv =
1
2
 cÖ¯’
=
1
2

x
2
=
x
4
cÖkœg‡Z, AvqZb = 512
 x 
x
2

x
4
= 512
 x3
= 2  4  512  x3
= 2  22
 29
 x3
= 2  22
 29
 x = (212
)
1
3  x = 24
 x = 16 m
15. The length of a cold storage is double its breadth. Its
height is 3 metres. The area of its four walls (including
the doors) is 108 m2
. Find its volume. (†Kvb wngvMv‡ii ˆ`N©¨
cÖ‡¯’i wØ¸Y I D”PZv 3 m| ìRvmn Gi Pvi †`qv‡ji †gvU
†ÿÎdj 108 m2
| wngvMv‡ii AvqZb wbY©q Kiæb|)
a 215 m3
b 216 m3
c 217 m3
d 218 m3
b
 mgvavb : awi, cÖ¯’ = x; ˆ`N©¨ = 2x
Pvi †`qv‡ji †ÿÎdj = 2  D”PZv  (ˆ`N©¨ + cÖ¯’)
= 2  3  (2x + x) = 18x
cÖkœg‡Z, 18x = 108 ev, x = 6
AvqZb = 2x  x  3 = 2  6  6  3 m3
= 216 m3
16. The length of hall is 20 metres and the width is 16
metres. The sum of the areas of the floor and roof is
equal to the sum of the areas of the four walls. Find the
volume of the hall. (GKwU K‡ÿi ˆ`N©¨ I cÖ¯’ h_vµ‡g 20 m I
16 m| †g‡S I Qv‡ì †ÿÎd‡ji mgwó Pvi †`qv‡ji †gvU
†ÿÎd‡ji mgvb| K‡ÿi AvqZb wbY©q Kiæb|)
a 2844.4 m3
b 2866.8 m3
c 2877.8 m3
d 2899.8 m2
a
 mgvavb : †g‡S I Qv‡ì †ÿÎd‡ji mgwó = 2  ˆ`N©¨  cÖ¯’
= 2  20  16 m2
= 640 m3
Pvi †`qv‡ji †ÿÎdj = 2  D”PZv  (ˆ`N©¨ + cÖ¯’)
= 2  D”PZv  (20 + 16)
= 72  D”PZv
cÖkœg‡Z, 72  D”PZv = 640
 D”PZv =
80
9
m
 AvqZb = 20  16 
80
9
m3
= 2844.44 m3
17. If V be the volume and S be the surface are of a cuboid
of dimensions a, b, c, then
1
V
is equal to (†Kvb Nb‡ÿ‡Îi
avi¸‡jv a, b, c Ges AvqZb I mgMÖ c„‡ôi †ÿÎdj h_vµ‡g V I
S n‡j
1
V
mgvbÑ)
a
S
2
(a + b + c) b
2
S 


1
a
+
1
b
+
1
c
c
25
a + b + c
d 2S (a + b + c) b
 mgvavb : AvqZb, V = abc ... ... (i)
mgMÖ c„‡ôi †ÿÎdj, S = 2(ab + bc + ca) ... ... (ii)
(ii)  (i) 
S
V
=
2(ab + bc + ca)
abc

S
V
= 2



1
c
+
1
a
+
1
b

1
V
=
2
S 


1
a
+
1
b
+
1
c
18. *The volume of a rectangular block of stone is 10368
dm3
. Its dimensions are in the ratio of 3 : 2 : 1. If its
entire surface is polished at 2 paise per dm2
, then the
total cost will be (AvqZvKvi GKwU cv_y‡i eø‡Ki AvqZb 10368
dm3
| Gi avi¸‡jvi AbycvZ h_vµ‡g 3 : 2 : 1| cÖwZ dm3
†ÿÎd‡j 2 cqmv Li‡P mgMÖ c„ô cwjk Ki‡Z †gvU LiPÑ)
a 31.50 Tk. b 31.68 Tk. c 63 Tk. d 63.36 Tk. d
 mgvavb : awi, ˆ`N©¨ 3x; cÖ¯’ 2x; D”PZv x
 AvqZb = 3x  2x  x = 6x3
cÖkœg‡Z, 6x3
= 10368
 x3
= 1728  x3
= 123
 x = 12
mgMÖ c„‡ôi †ÿÎdj = 2  (3x . 2x + 2x . x + x . 3x)
= 22x2
= 22  122
dm2
= 3168 dm2
 †gvU LiP = 3168  2 = 6336 cqmv
=
6336
100
UvKv [ 1 UvKv = 100 cqmv]
= 63.36 UvKv
19. The dimensions of a rectangular box are in the ratio 2 :
3 : 4 and the difference between the cost of covering it
with sheet of paper at the rate of 8 Tk. and 9.50 Tk. per
square metre is 1248 Tk. Find the dimensions of the
box in meters. (AvqZvKvi GKwU ev‡·i avi¸‡jvi AbycvZ 2 : 3
: 4| Gi c„‡ôi cÖwZ eM©wgUvi 8 UvKv I 9.50 UvKv nv‡i mgMÖ c„ô
KvMR w`‡q †gvov‡Z Li‡Pi eëavb 1248 UvKv| wgUvi GK‡Î
avi¸‡jv wbY©q Kiæb|) [www.examveda.com]
a 2 m, 12 m, 8 m b 4 m, 9 m, 16 m
c 8 m, 12 m, 16 m d None of these c
 mgvavb : awi, avi¸‡jv 2x, 3x, 4x|
mgMÖ c„‡ôi †ÿÎdj = 2  (2x . 3x + 3x.4x + 4x.2x) = 52x2
Li‡Pi cv_©K¨ = 9.5  52x2
– 8  52x2
= 78x2
cÖkœg‡Z, 78x2
= 1248
 x2
= 16  x = 4
avi¸‡jv : 2  4 = 8 m
3  4 = 12 m
4  4 = 16 m
20. *It is required to construct a big rectangular hall to
accommodate 500 persons, allowing 22.5 m3
space per
person. The height of the hall is to be kept at 7.5m,
while the total inner surface area of the walls must be
1200 sq. m. Then the length and breadth of the hall
respectively are (cÖ‡Z¨K e¨w³i Rb¨ 22.5 m3
AvqZb RvqMv
eiv× †i‡L 500 Rb e¨w³i Rb¨ GKwU eo AvqZvKvi Kÿ wbg©vY
Ki‡Z n‡e| K‡ÿi D”PZv 7.5 m Ges Pvi †`qv‡ji †gvU †ÿÎdj
1200 m2
ivL‡Z n‡e| ˆ`N©¨ I cÖ¯’ h_vµ‡gÑ)
a 40 m and 30 m b 45 m and 35 m
c 50 m and 30 m d 60 m and 20 m c
 mgvavb : AvqZb = 500  22.5 m3
= 11250 m3
ˆ`N©¨ = l; cÖ¯’ = b; D”PZv, h = 7.5 m
AvqZb = lbh
 lbh = 11250
 lb  7.5 = 11250
 lb = 1500

Pvi †`qv‡ji †ÿÎdj = 2h (l + b)
 2h (l + b) = 1200
 l + b =
1200
2  7.5
 l + b = 80 ... ... (i)
l – b = (l + b)2
– 4lb
= 802
– 4  1500
 l – b = 20 ... ... (ii)
(i) + (ii) 
l + b = 80
l – b = 20
2l = 100
 l = 50 m
(i) – (ii) 
l + b = 80
l – b = 20
(–) (+)
2b = 60
b = 30 m
21. A cuboidal water tank contains 216 litres of water. Its
depth is
1
3
of its length and breadth is
1
2
of
1
3
of the
difference between length and depth. The length of the
tank is (Nb‡ÿÎvKvi GKwU cvwbi U¨vs‡Ki aviYÿgZv 216
wjUvi| Gi MfxiZv ˆ`‡N©ï
1
3
Ask| cÖ¯’ ˆ`N©¨ I D”PZvi cv_©‡Kï
1
3
As‡ki
1
2
Ask| U¨vs‡Ki ˆ`N©¨Ñ) [www.examveda.com]
a 2 dm b 6 dm c 18 dm d 72 dm c
 mgvavb : awi, ˆ`N©¨ x; MfxiZv
x
3
;
cÖ¯’ =
1
2

1
3




x –
x
3
=
x
9
 AvqZb = x 
x
3

x
9
=
x3
27
cÖkœg‡Z,
x3
27
= 216 
1
1000
[ 1000 litres = 1 m3
]
 x3
=
216  27
1000
 x3
=
63
 33
103  x =
6  3
10
= 1.8 m
 x = 1.8  10 dm [ 1m = 10 dm]
 x = 18 dm|
22. The length of the longest rod that can be placed in a
room of dimensions 10 m  10 m  5 m is (10 m  10 m
 5 m mvB‡Ri GKwU K‡ÿ m‡e©v”P KZ ˆ`‡N©ï iW ivLv hv‡e?)
[www.examveda.com]
a 15 3 b 15 c 10 2 d 5 3 b
 mgvavb : Nb‡ÿ‡Îi wewfbœ kx‡l©i g‡a¨ m‡e©v”P `~iZ¡ K‡Y©i|
K‡Y©i ˆ`N©¨ = 102
+ 102
+ 52
m
= 15 m
23. Find the length of the longest rod that can be placed in
a room 16 m long, 12 m broad and 10
2
3
m high. (16 m
`xN©, 12 m cÖk¯Í I 10
2
3
m DuPz GKwU K‡ÿ m‡e©v”P KZ ˆ`‡N©ï iW
ivLv hv‡e?)
a 22
1
3
m b 22
2
3
m c 23 m d 68 m b
 mgvavb : i‡Wi ˆ`N©¨ = K‡Y©i ˆ`N©¨ = 162
+ 122
+



10
2
3
2
m
= 22
2
3
m
24. The volume of a rectangular solid is 210 cm3
and the
surface area is 214 cm2
. If the area of the base is 42
cm2
, then the edges of the rectangular solid are
(AvqZvKvi Nbe¯‘i AvqZb I c„‡ôi †ÿÎdj h_vµ‡g 210 cm3
I
214 cm2
| f‚wgi †ÿÎdj 42 cm2
n‡j avi¸‡jv KZ?)
a 3, 4, and 5 cm b 4, 5, and 6 cm
c 5, 6, and 7 cm d 6, 6, and 8 cm c
 mgvavb : awi, avi¸‡jv h_vµ‡g : l , b, h
AvqZb = lbh
c„‡ôi †ÿÎdj = 2 (lb + bh + lh)
f‚wgi †ÿÎdj = lb
kZ©g‡Z, lbh = 210
2(lb + bh + hl) = 214
lb = 42
h =
210
lb
 h =
210
42
 h = 5 cm
2 (42 + 5b + 5l) = 214
 42 + 5(b + l) = 107
 5 (b + l) = 65
l + b = 13 ... ... (i)
l – b = (l + b)2
– 4lb
= 132
– 4  42
 l – b = 1 ... ... (ii)
(i) + (ii) 
l + b = 13
l – b = 1
2l = 14
 l = 7
(i) – (ii) 
l + 6 = 13
l – b = 1
2b = 12
l = 7 cm
b = 6 cm
h = 5 cm
25. *How many bricks, each measuring 25 cm  11.25 cm 
6 cm, will be needed to build a wall 8 m  6 m  22.5
cm? (8 m  6m  22.5 cm gv‡ci GKwU †`qvj wbg©v‡Y 25 cm 
11.25 cm  6 cm gv‡ci KZ¸‡jv BU jvM‡e?)
a 5600 b 600 c 6400 d 7200 c
 mgvavb : B‡Ui AvqZb = 25 cm  11.25 cm  6 cm
= 1687.5 cm3
†`qv‡ji AvqZb = 8m  6 m  22.5 cm
= (8  100 cm)  (6  100 cm)  22.5 cm
= 10800000 cm3
 B‡Ui msL¨v =
10800000
1687.5
= 6400
26. The number of bricks, each measuring 25 cm  12.5 cm
 7.5 cm, required to construct a wall 6 m long, 5 m
high and 0.5 m thick, while the mortar occupies 5% of
the volume of the wall, is (6 m `xN©, 5m DuPz I 0.5 m cyiæ
GKwU †`qvj wbg©v‡Y 25 cm  12.5 cm  7.5 cm gv‡ci KZ¸‡jv
BU jvM‡e hw` †`qv‡ji AvqZ‡bi 5% c‡j¯Ív _v‡K?)
a 3040 b 5740 c 6080 d 8120 c

 mgvavb : B‡Ui AvqZb = 25 cm  12.5 cm  7.5 cm
=
25
100
m 
12.5
100
m 
7.5
100
m =
3
1280
m3
†`qv‡ji AvqZb = 6 m  5 m  0.5 m = 15 m3
†`qv‡j e¨eüZ †gvU B‡Ui AvqZb = 15 m3
Gi (100 – 5)%
=
95
100
 15 m3
=
57
4
m3
 B‡Ui msL¨v =
57
4
3
1280
= 6080
27. *50 men took a dip in a water tank 40 m long and 20 m
broad on a religious day. If the average displacement of
water by a man is 4 m3
, then the rise in the water level
in the tank will be : (agx©q Kvi‡Y 50 Rb †jvK 40 m `xN© I
20 m cÖk¯Í GKwU cyKz‡i Wze w`j| M‡o GKR‡bi Rb¨ hw` 4m3
cvwb AcmvwiZ nq, Z‡e cvwbi †j‡fj †gvU KZUzKz ev‡P?)
a 20 cm b 25 cm c 35 cm d 50 cm b
 mgvavb : awi,
cyKz‡ii MfxiZv = h m
 cyKz‡ii AvqZb = 40  20  h m3
= 800 h m3
cvwbi †j‡fj x m e„w× †c‡j
cyKz‡ii AvqZb = 40  20  (h + x) m3
= 800 (h + x) m3
 AvqZb e„w× ev cvwbi †gvU AcmviY
= [800 (x + h) – 800 h] m3
= 800 x m3
kZ©g‡Z, 800 x = 50  4
 x =
50  4
800
 x =
1
4
m
 x =
1
4
 100 cm = 25 cm
28. *A swimming bath is 24 m long and 15 m broad. When
a number of men dive into the bath, the height of the
water rise by 1 cm. If the average amount of water
displaced by one of the men be 0.1 cu. m, how many
men are there in the bath? (24 m `xN© I 15 m PIov GKwU
myBwgs K‡ÿ K‡qKRb †jvK Wze w`‡j cvwbi ¯Íi 1 cm ev‡o| hw`
M‡o GKR‡bi Rb¨ 0.1 m3
cwigvY cvwb AcmvwiZ nq, Z‡e KZRb
†jvK Wze w`‡qwQj?) [www.examveda.com]
a 32 b 36 c 42 d 46 b
 mgvavb : †gvU AcmvwiZ cvwbi AvqZb = 24 m  15 m  1 cm
= 24 m  15 m 
1
100
m
=
18
5
m3
 †gvU †jvK =
†gvU AcmvwiZ cvwbi AvqZb
M‡o GKR‡bi Rb¨ AcmvwiZ cvwbi AvqZb
=
18
5
0.1
= 36
29. *A school room is to be built to accommodate 70
children so as to allow 2.2 m2
of floor and 11 m3
of
space for each child. If the room be 14 metres long,
what must be its breadth and height? (70wU wkïi Rb¨
GKwU ¯‹zj Kÿ wbg©vY Ki‡Z n‡e †hb cÖ‡Z¨K wkïi Rb¨ †g‡Si
†ÿÎdj 2.2 m2
nq I 11 m3
¯’vb eivÏ _v‡K| K‡ÿi ˆ`N©¨ 14 m
n‡j cÖ¯’ I D”PZv KZ?)
[www.examveda.com]
a 11 m, 4 m b 11 m, 5 m
c 12 m, 5.5 m d 13 m, 6m b
 mgvavb : K‡ÿi AvqZb = 70  11 m3
= 770 m3
ˆ`N©¨, l = 14 m
awi, cÖ¯’ I D”PZv h_vµ‡g b I h|
 lbh = 770
 14  bh = 770
 bh = 55
†g‡Si †ÿÎdj lb = 2.2  70
 14  b = 2.2  70
 b = 11 m
bh = 55
 h =
55
11
 h = 5m
30. A rectangular tank measuring 5 m  4.5 m  2.1 m is
dug in the centre of the field measuring 13.5 m by 2.5
m. The earth dug out is evenly spread over the
remaining portion of the field. How much is the level of
the field raised? (13.5 m  2.5 m gv‡ci GKwU gv‡Vi gvSLv‡b
5 m  4.5 m  2.1 m gv‡ci GKwU cyKzi Lbb Kiv n‡jv| LbbK…Z
gvwU m¤ú~Y© gv‡V mgvbfv‡e Qwo‡q w`‡j gvV KZUzKz DuPz n‡e?)
a 4 m b 4.1 m c 4.2 m d 4.3 m c
 mgvavb : gv‡Vi †ÿÎdj = 13.5  2.5 m2
= 33.75 m2
cyKz‡ii DcwiZ‡ji †ÿÎdj = 5m  4.5 m = 22.5 m2
cyKyi ev‡` gv‡Vi †ÿÎdj = (33.75 – 22.5) m2
= 11.25 m2
LbbK…Z gvwUi cwigvY = cyKz‡ii AvqZb = 5  4.5  2.1 m3
= 47.25 m3
 gv‡Vi D”PZv e„w× =
gvwUi AvqZb
gvwU Øviv Av”Qvw`Z As‡ki †ÿÎdj
=
47.25 m3
11.25 m2 = 4.2 m
31. A plot of land in the form of a rectangle has dimensions
240 m  180 m. A drainlet 10 m wide is dug all around
it (outside) and the earth dug out is evenly spread over
the plot, increasing its surace level by 25 cm. The depth
of the drainlet is. (240 m  180 m gv‡ci AvqZvKvi GKLÐ
Rwgi Pvicv‡k 10 m PIov bvjx Lbb Kiv n‡jv Ges LbbK…Z gvwU
Rwgi me©Î mgfv‡e Qov‡bv n‡jv| gv‡Vi †j‡fj/D”PZv 25 cm
evoj| bvjxi MfxiZv KZ?) [www.examveda.com]
a 1.223 m b 1.225 m c 1.227 m d 1.229 m c
 mgvavb : gv‡Vi †ÿÎdj = 240  180 m2
= 43200 m2
bvjxmn gv‡Vi †ÿÎdj = (240 + 2  10)  (180 + 2  10) m2
= 52000 m2
 bvjxi †ÿÎdj = [52000 – 43200] m2
= 8800 m2
LbbK…Z gvwUi AvqZb = 43200 m2
 25 cm
= 43200 
25
100
m3
= 10800 m3
bvjxi MfxiZv =
10800
8800
m = 1.227 m
32. *A cistern, open at the top, is to be lined with sheet of
lead which weights 27 kg/m2
. The cistern is 4.5 m long
and 3 m wide and holds 50 m3
. The weight of lead
required is (gyL †Lvjv GKwU †PŠev”Pvi †fZ‡ii Ask mxmv w`‡q
Ave„Z Ki‡Z n‡e| cÖwZ 1 m2
mxmvi IRb 27 kg| †PŠev”Pvi ˆ`N©¨,
cÖ¯’ I †fZ‡ii As‡ki AvqZb h_vµ‡g 4.5 m, 3m I 50 m3
| GB
Kv‡R cÖ‡qvRbxq mxmvi IRbÑ) [www.examveda.com]
a 1660.5 kg b 1764.5 kg
c 1860.5 kg d 1864.5 kg d

 mgvavb : †PŠev”Pvi †g‡Si †ÿÎdj = 4.5  3 m2
= 13.5 m2
 MfxiZv =
50
13.5
m =
100
27
m
l = 4.5 m, b = 3m, h =
100
27
m
†fZ‡ii As‡ki †gvU †ÿÎdj = lb + 2h (l + 6)
=



13.5 + 2 
100
27
 (4.5 + 3) m2
=
1243
18
m2
 cÖ‡qvRbxq mxmvi cwigvY =
1243
18
 27 kg = 1864.5 kg
33. If a river 2.5 m deep and 45 m wide is flowing at the
rate of 3.6 km per hour then the amount of water that
runs into the sea per minute is (2.5 m Mfxi I 45 m PIov
GKwU b`x NÈvq 3.6 km †e‡M cÖevngvb| cÖwZ wgwb‡U wK cwigvY
cvwb mvM‡i cwZZ n‡”Q?)
a 6650 cu. m b 6750 cu. m
c 6850 cu. m d 6950 cu. m b
 mgvavb : 1 wgwb‡U cvwb AwZµg K‡i =
3.6
60
km
=
3.6  1000
60
m
= 60 m
 1 wgwb‡U cÖevwnZ cvwb ev mvM‡i cwZZ cvwbi
cwigvY = 60 m  45 m  2.5 m = 6750 m3
34. *A rectangular water tank is 80 m  40 m. Water flows
into it through a pipe 40 sq. cm at the opening at a
speed of 10 km/ hr. By how much, the water level will
rise in the tank in half an hour? (40 cm2
cÖ¯’‡”Q` wewkó
GKwU cvB‡ci ga¨ w`‡q 10 km/hr †e‡M cvwb 80 m  40 m
gv‡ci GKwU U¨vs‡K cÖevwnZ n‡”Q| AvaNÈv ci U¨vs‡K cvwbi ¯Íi
KZUzKz n‡e?)
a
3
2
cm b
4
9
cm
c
5
8
cm d None of these c
 mgvavb : cvB‡ci cÖ¯’‡”Q‡`i †ÿÎdj = 40 cm2
=
40
100  100
m2
=
1
250
m2
cÖevn †eM = 10 km/hr
= 10  1000 m/hr
= 10000 m/hr
 cÖwZ NÈvq cÖevwnZ cvwbi cwigvY = cvB‡ci cÖ¯’‡”Q`  cÖevn †eM
=
1
250
 10000 m3
= 40 m3
 Avav NÈv ev 0.5 NÈvq cÖevwnZ cvwbi cwigvY = 0.5  40 m3
= 20 m3
cvwbi ¯Íi h m n‡j 80  40  h = 20
 h =
20
80  40
m  h =
1
160
 100 cm =
5
8
cm
35. A rectangular tank is 225 m by 162 m at the base. With
what speed must water flow into it through an aperture
60 cm by 45 cm so that the level may be raised 20 cm in
5 hours? (AvqZvKvi GKwU U¨vsK Gi f‚wgi gvc 225 m  162
m| 60 cm  45 cm cÖ¯’‡”Q`wewkó GKwU wQ‡`ªi ga¨ w`‡q KZ
†e‡M U¨vs‡K cvwb cÖ‡ek Ki‡j 5 NÈv ci cvwbi ¯Íi 20 cm e„w×
cv‡e?) [www.examveda.com]
a 5000 m/hr b 5200 m/hr
c 5400 m/hr d 5600 m/hr c
 mgvavb : 5 NÈvq cÖevwnZ cvwbi AvqZb
= 225 m  162 m  20 cm
= 225 m  162 m 
20
100
m
= 7290 m3
1 NÈvq cÖevwnZ cvwbi AvqZb =
7290
5
m3
= 1458 m3
cvB‡ci cÖ¯’‡”Q` = 60 cm  45 cm
=
60
100
m 
45
100
m = 0.27 m2
NÈvq cÖevwnZ cvwbi cwigvY = cvB‡ci cÖ¯’‡”Q`  cÖevn †eM
 1458 = 0.27  cÖevn †eM
 cÖevn †eM = 5400 m/hr
36. The water in a rectangular reservoir having a base 80
m by 60 m is 6.5 m deep. In what time can the water be
emptied by a pipe of which the cross-section is a square
of side 20 cm, if the water runs through the pipe at the
rate of 15 km per hour? (80 m  60 m gv‡ci f‚wgwewkó
GKwU AvqZvKvi Rjvav‡i 6.5 m Mfxi cvwb Av‡Q| 20 cm
aviwewkó eM©vKvi wQ`ªhy³ cvB‡ci mvnv‡h¨ 15 km/hr cÖevn‡e‡M
KZ mg‡q RjvaviwU Lvwj Kiv hv‡e?)
a 26 hrs b 42 hrs c 52 hrs d 65 hrs c
 mgvavb : cvwbi cwigvY = 80  60  6.5 m3
= 31200 m3
cvB‡ci cÖ¯’‡”Q` = (20 cm)2
=



20
100
m
2
=
1
25
m2
NÈvq cÖevwnZ cvwbi cwigvb = cvB‡ci cÖ¯’‡”Q`  cÖevn †eM
=
1
25
 (15  1000) m3
= 600 m3
cÖ‡qvRbxq mgq =
31200
600
= 52 hrs
37. Rita and Meeta both are having lunch boxes of a
cuboidal shape. Length and breadth of Rita's lunch
box are 10% more than that of Meeta's lunch box, but
the depth of Rita's lunch box is 20% less than that of
Meeta's lunch box. The ratio of the capacity of Rita's
lunch box to that of Meeta's lunch box is (wiZv I wgZv
Df‡qi Kv‡QB NbvKvi jvÂe· Av‡Q| wiZvi e‡·i ˆ`N©¨ I cÖ¯’
wgZvi e‡·i ˆ`N©¨ I cÖ¯’ A‡cÿv 10% †ewk wKš‘ wiZvi e‡·i
MfxiZv wgZvi e‡·i MfxiZvi Zzjbvq 20% Kg| wiZv I wgZvi
jvÂe‡·i aviYÿgZvi AbycvZÑ) [www.examveda.com]
a 11 : 15 b 15 : 11 c 121 : 125 d 125 : 121 c
 mgvavb : wgZvi Zzjbvq wiZvi e‡·i ˆ`N©¨, x% = 10% [†ewk]
cÖ¯’, Y% = 10% [†ewk]
MfxiZv, Z% = – 20% [Kg]

wiZvi e‡·i aviYÿgZv
wgZvi e‡·i aviYÿgZv =



1 +
x
100 


1 +
y
100 


1 +
z
100
=



1 +
10
100 


1 +
10
100 


1 –
10
100
=
11
10

11
10

4
5
=
121
125
38. The sum of the length, breadth and depth of a cuboid is
19 cm and its diagonal is 5 5 cm. It surface area is
(GKwU Nb‡ÿ‡Îi ˆ`N©¨, cÖ¯’ I MfxiZvi mgwó 19 cm I K‡Y©i
ˆ`N©¨ 5 5 cm| mgMÖ c„‡ôi †ÿÎdjÑ
a 125 cm2
b 236 cm2
c 361 cm2
d 486 cm2
b

 mgvavb : awi, ˆ`N©¨, cÖ¯’ I MfxiZv h_vµ‡g l, b, h
kZ©g‡Z, l + b + h = 19
l2
+ b2
+ h2
= 5 5
 l2
+ b2
+ h2
= 125
(l + b + h)2
= l2
+ b2
+ h2
+ 2 (lb + bh + hl)
 192
= 125 + c„‡ôi †ÿÎdj
 c„‡ôi †ÿÎdj = (361 – 125) cm2
= 236 cm2
39. *The sum of perimeters of the six faces of a cuboid is 72
cm and the total surface area of the cuboid is 16 cm2
.
Find the longest possible length that can be kept inside
the cuboid. (GKwU Nb‡ÿ‡Îi 6wU Z‡ji cwimxgvi mgwó 72 cm
I mgMÖ c„‡ôi †ÿÎdj 16 cm2
| Nb‡ÿÎwUi Af¨šÍ‡i †h‡Kvb `yB
we›`yi g‡a¨Kvi m¤¢ve¨ m‡e©v”P `~iZ¡ KZ?) [www.examveda.com]
a 5.2 cm b 7.8 cm c 8.05 cm d 8.36 cm c
 mgvavb : awi, Nb‡ÿÎwUi avi¸‡jv a, b, c
 †gvU cwimxgv = 8 (a + b + c)
c„‡ôi †ÿÎdj = 2(ab + bc + ca)
KY©, d = a2
+ b2
+ c2
kZ©g‡Z, 8(a + b + c) = 72
 a + b + c = 9
2(ab + bc + ca) = 16
(a + b + c)2
= a2
+ b2
+ c2
+ 2(ab + bc + ca)
 a2
= d2
+ 16
 d2
= 81 – 16
 d2
= 65
 d = 65
 d = 8.06 cm
40. A swimming pool 9 m wide and 12 m long is 1 m deep
on the shallow side and 4 m deep on the deeper side. Its
volume is (9 m PIov I 12m `xN© GKwU myBwgs cy‡ji AMfxi
cv‡ki MfxiZv 1 m I MfxiZg cv‡ki MfxiZv 4m myBwgs cy‡ji
AvqZb|) [www.examveda.com]
a 208 m3
b 270 m3
c 360 m3
d 408 m3
b
 mgvavb : ˆ`N©¨ I MfxiZvi Av‡jv‡K wPÎ
A¼b Kwi :
myBwgs cy‡ji †ÿÎdj (cÖ¯’ eivei)
12 m
4 m
1m
A B
D C
O
= OABC UªvwcwRqvg Gi †ÿÎdj
=
1
2
 (OA + BC)  AB
=
1
2
 (1 + 4)  12 m2
=
1
2
 5  12 m2
= 30 m2
 AvqZb = 30  9 m3
= 270 m3
41. *Length of a rectangular solid is increased by 10% and
breadth is decreased by 10%. Then the volume of the
solid. (GKwU AvqZvKvi Nbe¯‘i ˆ`N©¨ 10% e„w× I cÖ¯’ 10% n«vm
Ki‡j Gi AvqZbÑ)
a remains unchanged b decreased by 1%
c decreased by 10% d increases by 10% b
 mgvavb : ˆ`N©¨ e„w×, x% = 10%
cÖ¯’ e„w×, y% = – 10% [n«vm]
D”PZv e„w×, z% = 0 [AcwiewZ©Z]
 AvqZ‡bi kZKiv cwieZ©b
= x% + y% + z% +
xy + yz + zx
100
% +
xyz
100  100
%
= 10% – 10% + 0 +
10  (–10) + 0 + 0
100
%
+
10  (–10)  0
100  100
%
=
– 100
100
% = – 1% [n«vm]
42. The length, breadth and height of a cuboid are in the
ratio 1 : 2 : 3. The length, breadth and height of the
cuboid are increased by 100%, 200% and 200%
respectively. Then the increase in the volume of the
cuboid is. (GKwU Nb‡ÿ‡Îi ˆ`N©¨, cÖ¯’ I D”PZvi AbycvZ
h_vµ‡g 1 : 2 : 3| ˆ`N©¨, cÖ¯’ I D”PZv h_vµ‡g 100%, 200% I
200% e„w× Ki‡j AvqZb KZ¸Y e„w× cv‡e?)
a 5 times b 6 times
c 12 times d 17 times d
 mgvavb : ˆ`N©¨ e„w×, x% = 100%
cÖ¯’ e„w×, y% = 200%
D”PZv e„w×, z% = 200%
 AvqZb e„w× = (x + y + z)% +
xy + yz + zx
100
% +
xyz
100  100
= (100 + 200 + 200) % +
100  200 + 200  200 + 200  100
100
%
+
100  100  200
100  100
%
= 500 % + 800% + 400%
= 1700%
=
1700
100
¸Y = 17 ¸Y
43. A rectangular piece of cardboard 18 cm  24 cm is
made into an open box by cutting a square of 5 cm side
from each corner and building up the side. Find the
volume of the box in cu. cm. (18 cm  24 cm gv‡ci GKwU
AvqZvKvi KvW©‡evW© Gi cÖ‡Z¨K †KvYv †_‡K 5 cm aviwewkó GKwU
K‡i eM© †K‡U †bqv n‡jv| eM©¸‡jv w`‡q †`qvj evwb‡q KvW©‡evW©wU‡K
GKwU †Lvjv ev‡· cwiYZ Ki‡j Gi AvqZb n‡eÑ)
a 216 b 432
c 560 d None of these c
 mgvavb :
 e‡·i ˆ`N©¨, l = [24 – 2  5] cm
= 14 cm
cÖ¯’, b = [18 – 2  5] = 8cm
D”PZv, h = 5 cm
 AvqZb = lbh = 14  8  5 cm3
= 560 cm3
24 cm 5cm 18cm
5cm
44. An open box is made by cutting the congruent squares
from the corners of a rectangular sheet of cardboard of
dimensions 20 cm  15 cm. If the side of each square is
2 cm, the total outer surface area of box is (20 cm  15
cm gv‡ci GKwU AvqZvKvi KvW©‡evW© Gi cÖ‡Z¨K †KvYv †_‡K 2 cm
aviwewkó eM© †K‡U wb‡q GKwU †Lvjv ev· ˆZwi Kiv n‡jv| mgMÖ
c„‡ôi †ÿÎdjÑ) [www.examveda.com]
a 148 cm2
b 284 cm2
c 316 cm2
d 460 cm b

 mgvavb : ˆ`N©¨, l = [20 – 2  2] cm
= 16 cm
cÖ¯’, b = [15 – 2  2] cm
= 11 cm
MfxiZv, h = 2 cm
†Lvjv gy‡Li †ÿÎdj = lb
 c„‡ôi †ÿÎdj = 2(lb + bh + lh) – lb
= lb + 2h (b + l)
= [16  11 + 2  2  (16 + 11)] cm2
= 284 cm2
45. A closed box made of wood of uniform thickness has
length, breadth and height 12 cm, 10 cm and 8 cm
respectively. If the thickness of the wood is 1 cm, the
inner surface area is (1 cm mylgfv‡e cyiæ KvV w`‡q 12 cm 
10 cm  8 cm gv‡ci e× ev· ˆZwi Kiv n‡jv| ev‡·i Af¨šÍixY
c„‡ôi †gvU †ÿÎdjÑ) [www.examveda.com]
a 264 cm2
b 376 cm2
c 456 cm2
d 696 cm2
b
 mgvavb : †fZ‡ii As‡kiÑ
ˆ`N©¨, l = (12 – 2  1) cm = 10 cm
cÖ¯’, b = (10 – 2  1) = 8 cm
D”PZv, h = (8 – 2  1) cm = 6 cm
 wb‡Y©q †ÿÎdj = 2(lb + bh + lh)
= 2  [10  8 + 8  6 + 10  6] cm2
= 376 cm2
46. *A covered wooden box has the inner measures as 115
cm, 75 cm and 35 cm and the thickness of wood is 2.5
cm. Find the volume of the wood. (mylgfv‡e cyiæ e× Kv‡Vi
ev‡·i †fZ‡ii w`‡Ki gvc 115 cm  75 cm  35 cm Ges cyiæZ¡
2.5 cm| Kv‡Vi (ïaygvÎ) AvqZb KZ?) [www.examveda.com]
a 81000 cu. cm b 81775 cu. cm
c 82125 cu. cm. d None of these c
 mgvavb : evB‡ii As‡ki
ˆ`N©¨ = (115 + 2  2.5) cm = 120 cm
cÖ¯’ = (75 + 2  2.5) cm = 80 cm
D”PZv = (35 + 2  2.5) cm = 40 cm
 mgMÖ ev‡·i AvqZb = 120  80  40 cm3
= 384000 cm3
duvKv As‡ki (wfZ‡ii) AvqZb = 115  75  35 cm3
= 301875 cm3
 Kv‡Vi AvqZb = [384000 – 301875] cm3
= 82125 cm3
47. The dimensions of an open box are 52 cm  40 cm  29
cm. Its thickness is 2 cm. If 1 cu. cm of metal used in
the box weighs 0.5 gm, then the weight of the box is
(GKwU †Lvjv ev‡·i gvÎv¸‡jv 52 cm  40 cm  29 cm Ges GwU
2 cm cyiæ| hw` 1 Nb †m.wg. avZz e¨eüZ nZ Z‡e ev‡·i IRb nZ
0.5 gm| ev·wUi IRb KZ?) [www.examveda.com]
a 6.832 kg b 7.576kg c 7.76 kg d 8.56 kg a
 mgvavb : ev·wUi evB‡ii AvqZb = (52  40  29) cm3
= 60, 320 cm3
†h‡nZz ev·wUi 1 gyL †Lvjv Ges cyiæZ¡ 2 cm
myZivs, ev·wUi †fZ‡ii ˆ`N©¨ = 52 – (2 + 2) = 48 cm
,, cÖ¯’ = 40 – (2 + 2) = 36 cm
,, D”PZv = 29 –2 = 27 cm
A_©vr, ev·wUi †fZ‡ii AvqZb = (48  36  27) cm3
= 46, 656 cm3
myZivs, ev·wU‡Z e¨eüZ avZzi AvqZb = evB‡ii AvqZb –
†fZ‡ii AvqZb = (60, 320 – 46, 656) cm3
= 13, 664 cm3
 1 cm3
avZzi IRb 0.5 gm
 13, 664 cm3
avZzi IRb = 0.5  13664 gm
= 6, 832 gm
= 6.832 kg
48. An open box is made of wood 3 cm thick. Its external
dimensions are 1.46 m, 1.16 m and 8.3 dm. The cost of
painting the inner surface of the box at 50 paise per
100 sq. cm is (GKwU Kv‡Vi ˆZwi †Lvjv ev‡·i cyiæZ¡ 3 cm. GwUi
evwn¨K gvÎv¸‡jv nj h_vµ‡g 1.46 m, 1.16 m Ges 8.3 dm| 100
eM© †m.wg. is Ki‡Z hw` 50 cqmv LiP nq Zvn‡j ev‡·i Af¨šÍxY
Zjmg~n is Ki‡Z †gvU LiP n‡eÑ)
a 138.50 Tk. b 277 Tk. c 415.50 Tk. d 554 Tk. b
 mgvavb : evwn¨K gvÎv¸‡jv †m.wg.,
G wb‡j cvB 146 cm  116 cm  83 cm
†h‡nZz cyiæZ¡ 3 cm
ca
ca
bc
bc
ab
 Af¨šÍixY ˆ`N©¨, a = 146 – (3 + 3) = 140 cm
,, cÖ¯’, b = 116 – (3 + 3) = 110 cm
,, D”PZv, c = 83 – 3 = 80 cm
ev·wUi Dc‡ii ab Zj †Lvjv e‡j is Ki‡Z n‡e evwK 5wU
Zj| myZivs, is K…Z Z‡ji †ÿÎdj = ab + 2bc + 2ca
= 140  110 + 2  110  80 + 2  80  140
= 55, 400 cm2
100 cm2
is Ki‡Z LiP nq 50 cqmv
 55, 400 cm2
,, ,, ,, ,,
50  55

cqmv
= 27,700 cqmv
= 277 UvKv [ 1 UvKv = 100 cqmv]
49. *A cistern of capacity 8000 litres measures externally
3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick.
The thickness of the bottom is (8000 wjUvi aviYÿgZvi
GKwU †PŠev”Pvi evwn¨K gvÎv¸‡jv nj 3.3 m, 2.6 m I 1.1 m
†PŠev”Pvi †`qv‡ji cyiæZ¡ 5 cm wb¤œZ‡ji cyiæZ¡ njÑ)
a 90 cm b 1 dm c 1 m d 1.1 m b
 mgvavb : Avgiv Rvwb,
1000 wjUvi cvwbi AvqZb 1 m3
 8000 ,, ,, ,,
8000
1000
= 8 m3
awi, †PŠev”Pvi wb¤œZ‡ji cyiæZ¡ x wg.
†`qv‡ji cyiæZ¡ = 5 cm = 0.05 m
 †PŠev”Pvi Af¨šÍixb AvqZb,
= (3.3 – 2  0.05)  (2.6 – 2  0.05) + (1.1 – x)
= 3.2  2.5  (1.1 – x)
= 8(1.1 – x)
cÖkœg‡Z, 8(1.1 – x) = 8
1.1 – x) = 8
 x = 1.1 – 1 = 0.1 = 1 dm
50. If a metallic cuboid weighs 16 kg, how much would a
miniature cuboid of metal weigh, if all dimensions are
reduced to one-fourth of the original? (hw` GKwU avZe
Nb‡ÿ‡Îi IRb 16 kg nq, Z‡e Gi gvÎv¸‡jvi GK-PZz_©vsk
gvÎvwewkó Nb‡ÿ‡Îi IRb KZ n‡e?0
a 0.25 kg b 0.50 kg c 0.75 kg d 1 kg a
 mgvavb : awi, eo Nb‡ÿ‡Îi gvÎv¸‡jv nj a, b I c
 †QvU Nb‡ÿ‡Îi gvÎv¸‡jv n‡e
a
4
,
b
4
I
c
4
eo Nb‡ÿ‡Îi IRb 16 kg
awi, †QvU ,, ,, w

abc
a
4

b
4

c
4
=
16
w
 w =
16
64
=
1
4
kg = 0.25 kg

51. *A rectangular water tank is open at the top. Its
capacity is 24 m3
. Its length and breadth are 4 m and 3
m respectively. Ignoring the thickness of the material
used for building the tank, the total cost of painting the
inner and outer surfaces of the tank at the rate of 10
Tk. per m2
is (GKwU AvqZvKvi U¨vs‡Ki Dc‡ii gyL †Lvjv| Gi
aviYÿgZv 24 m3
| Gi ˆ`N©¨ I cÖ¯’ h_vµ‡g 4 m I 3 m|
U¨vsKwUi MVb Dcv`v‡bi cyiæZ¡ we‡ePbv bv K‡i cÖwZ eM©wgUvi 10
UvKv nv‡i U¨vsKwUi †fZi I evB‡ii c„ômg~n is Ki‡Z †gvU LiP
n‡eÑ) [www.examveda.com]
a 400 Tk. b 500 Tk. c 600 Tk. d 800 Tk. d
 mgvavb : U¨vsKwUi MVb Dcv`v‡bi cyiæZ¡ we‡ePbv bv Ki‡j Gi
evwn¨K I Af¨šÍixY gvÎvmg~n mgvb n‡e d‡j U¨vswUi †fZ‡ii c„ô
is Ki‡Z hZ LiP n‡e evB‡ii c„ô is Ki‡ZI GKB LiP n‡e|
U¨vsKwUi †fZ‡ii ˆ`N©¨ a = 4 m
,, cÖ¯’ b = 3m
U¨vsKwUi D”PZv c n‡j, abc = 24
 c =
24
ab
=
24
4  3
= 2m
U¨vsKwUi Dc‡ii c„ô †Lvjv e‡j †gvU 5wU c„ô is Ki‡Z n‡e|
c„ô¸‡jv n‡jv 1wU ab Zj, 2wU bc Zj I 2wU ca Zj
†gvU is K…Z c„‡ôi †ÿÎdj = ab + 2bc + 2ca
= 4  3 + 2  3  2 + 2  2  4
= 40 m2
1 m2
is Ki‡Z LiP n‡e 10 UvKv
 40 m2
,, ,, ,, ,, 10  40 UvKv
= 400 UvKv
myZivs †fZ‡ii c„ô is Ki‡Z 400 UvKv n‡j evB‡ii c„ô is
Ki‡ZI LiP n‡e 400 UvKv †gvU LiP 400 + 400 = 800 UvKv|
52. If the areas of three adjacent faces of a cuboid are x, y,
z respectively, then the volume of the cuboid is (hw`
GKwU NbKvK…wZi e¯‘i msjMœ wZbwU c„‡ôi †ÿÎdj h_vµ‡g x, y I
z nq Z‡e e¯‘wUi AvqZb n‡eÑ) [www.examveda.com]
a xyz b 2xyz c xyz d 3 xyz c
 mgvavb : awi, NbKvK…wZi e¯‘i gvÎv¸‡jv h_vµ‡g a, b I c
 AvqZb abc = ?
cÖkœg‡Z, ab = x, bc = y, ca = z
mgxKiY 3wU‡K ¸Y Ki‡j ab . bc. ca = xyz
 a2
b2
c2
= xyz  (abc)2
= xyz  abc = xyz
53. If the areas of the three adjacent faces of a cuboidal
box are 120 cm2
, 72 cm2
and 60 cm2
respectively, then
find the volume of the box. (hw` GKwU NbKvK…wZi ev‡·i
msjMœ wZbwU c„‡ôi †ÿÎdj h_vµ‡g 120 cm2
, 72 cm2
I 60 cm2
Zvn‡j ev·wUi AvqZbÑ)
a 720 cm3
b 864 cm3
c 7200 cm3
d (72)2
cm3
a
 mgvavb : awi, ev·wUi gvÎv¸‡jv h_vµ‡g a, b I c
 ev·wUi AvqZb abc = ?
cÖkœg‡Z, ab = 120, bc = 72, ca = 60
mgxKiY 3wU‡K ¸Y Ki‡j cvB,
ab . bc. ca = 120  72  60
 a2
b2
c2
= 518, 400
 abc = 518  400 = 720 cm3
54. If the areas of three adjacent faces of a rectangular
block are in the ratio of 2 : 3 : 4 and its volume is 9000
cu. cm.; then the length of the shortest side is (hw` GKwU
AvqZvKvi ev‡·i msjMœ 3wU c„‡ôi †ÿÎdj¸‡jvi AbycvZ 2 : 3 : 4
nq Ges Gi AvqZb 9000 Nb cm; Zvn‡j me‡P‡q †QvU evûi ˆ`N©¨
n‡eÑ) [www.examveda.com]
a 10 cm b 15 cm c 20 cm d 30 cm b
 mgvavb : ev·wUi gvÎv¸‡jv a, b I c n‡j
AvqZb abc = 9000 cm3
cÖkœg‡Z, ab : bc : ca = 2 : 3 : 4
 ab = 2x ... ... (i) i  ii  iii Ki‡j
bc = 3x ... ... (ii) ab. bc . ca = 2x  3x  4x
ca = 4x ... ... (iii) ev, (abc)2
= 24x3
abc = 9000.... (iv) ev, x3
=
(9000)2
24
ev, x = 150
(iv)  (i) K‡i,
abc
ab
=
9000
2x
 c =
9000
2  150
= 30 cm
(iv)  (ii) K‡i,
abc
bc
=
9000
3x
 a =
9000
3  150
= 20 cm
(iv)  (i) K‡i,
abc
ca
=
9000
4x
 b =
9000
4  150
= 15 cm
 me‡P‡q †QvU ˆ`N©¨ 15 cm
55. The dimensions of a certain machine are 48  30 
52. If the size of the machine is increased
proportionately until the sum of its dimensions equals
156, what will be the increase in the shortest side?
(GKwU †gwk‡bi gvÎv¸‡jv n‡jv 48  30  52 hw` †gwk‡bi
mvBR AvbycvwZKfv‡e e„w× Kiv nq hZÿY bv gvÎv¸‡jvi †hvMdj
156, me‡P‡q †QvU evûwU KZUzKz e„w× cv‡e?) [www.examveda.com]
a 4 b 6 c 8 d 9 b
 mgvavb : gvÎv¸‡jvi cÖv_wgK †hvMdj = 48 + 30 + 52 = 130
e„w×i cwigvY = 156 – 130 = 26
myZivs †QvU evûwU e„w× cv‡e =
30
130
 26 = 6
56. If a metal slab of size 1 m  20 cm  1 cm is melted to
another slab of 1 mm thickness and 1 m width, then the
length of the new slab thus formed will be (hw` 1m  20
cm  1 cm gvÎvwewkó GKwU avZe ¯ø¨ve‡K Mwj‡q 1 mm cyiæZ¡, 1
m cÖ¯’wewkó Av‡iKwU bZzb ¯ø¨v‡e cwiYZ Kiv nq Z‡e bZzb
¯ø¨vewUi ˆ`N©¨ n‡eÑ)
a 200 cm b 400 cm c 600 cm d 1000 cm a
 mgvavb : bZzb ¯ø¨v‡ei AvqZb = cyivZb ¯ø¨v‡ei AvqZb
ˆ`N©¨  cÖ¯’  cyiæZ¡ = 1 m  20 cm  1 cm
 ˆ`N©¨  1 m  1 mm = 1 m  20 cm  10 mm
 ˆ`N©¨ = 200 cm
57. Rahul hired a contractor to dig a well of 10 metres
length, 10 metres breadth and 10 metres depth for
40000. However, when the contractor was about to
start the work, he changed his mind and asked him to
get two wells dug, each with a length of 5 metres,
breadth of 5 metres and depth of 5 metres. How much
should Rahul pay to the contractor? (ivûj 40000 UvKv
w`‡q 10 m ˆ`N©¨ 10 m cÖk¯Í I 10 m MfxiZvwewkó GKwU K…c
Lb‡bi Rb¨ GKRb wVKv`vi wb‡qvM w`‡jb| wVKv`vi hLb KvR ïiæ
Ki‡eb ZLb ivûj Zuvi gb cwieZ©b Ki‡jb Ges wVKv`vi‡K ej‡jb
5 m ˆ`N©¨, 5 wgUvi cÖk¯Í I 5 m MfxiZvwewkó 2wU K‚c Lbb
Ki‡Z| wVKv`vi‡K Zuvi KZ UvKv w`‡Z n‡e?)
a 1000 Tk. b 2000 Tk.
c 4000 Tk. d None of these a

 mgvavb : 2wU K‚c Lbb Ki‡Z LiP x UvKv n‡j
eo K‚‡ci AvqZb
†QvU 2wU K‚‡ci AvqZb =
eo K‚‡ci LiP
†QvU 2wU K‚‡ci LiP

10  10  10
2  5  5  5
=
40000
x
 x =
2  5  5  5  40000
10  10  10
= 10000 UvKv
58. Each side of a cube measures 8 metres. What is the
volume of the cube? (GKwU Nb‡Ki cÖwZwU evûi ˆ`N©¨ 8 wgUvi|
NbKwUi AvqZb KZ?) [www.examveda.com]
a 72 cu. m b 144 cu. m
c 196 cu. m d None of these d
 mgvavb : Avgiv Rvwb, Nb‡Ki AvqZb = a3
GLv‡b, evûi ˆ`N©¨ a = 8 m
 AvqZb = a3
= (8)3
= 512 m3
59. The perimeter of one face of a cube is 20 cm. Its
volume must be (Nb‡Ki GKwU c„‡ôi cwimxgv 20 cm| GwUi
AvqZb n‡eÑ)
a 125 cm3
b 400 cm3
c 1000 cm3
d 8000 cm3
a
 mgvavb : Nb‡Ki GKwU evûi ˆ`N©¨ a n‡j Gi †h‡Kvb c„‡ôi cwimxgv 4a
cÖkœg‡Z, 4a = 20  a = 5 cm
myZivs NbKwUi AvqZb = a3
= 53
= 125 cm3
60. Total surface area of a cube whose side is 0.5 cm is (0.5
cm evûwewkó GKwU Nb‡Ki mgMÖZ‡ji †ÿÎdj n‡eÑ)
[www.examveda.com]
a
1
4
cm2
b
1
8
cm2
c
3
4
cm2
d
3
2
cm2
d
 mgvavb : Nb‡Ki 6wU Zj we`¨gvb, Nb‡Ki evûi ˆ`N©¨ a n‡j,
cÖwZwU Z‡ji †ÿÎdj a2
 mgMÖZ‡ji †ÿÎdj = 6a2
= 6  (0.5)2
=
3
2
cm2
61. *The cost of the paint is 36.50 Tk. per kg. If 1 kg of
paint covers 16 square feet, how much will it cost to
paint outside of a cube having 8 feet each side? (cÖwZ
†KwR i‡Oi g~j¨ 36.50 UvKv| hw` 1 kg iO w`‡q 16 eM©dzU iO Kiv
hvq Z‡e 8 dzU evûwewkó GKwU Nb‡Ki evwn¨K c„ô iO Ki‡Z KZ
LiP n‡e?)
a 692 Tk. b 768 Tk. c 876 Tk. d 972 TK.
e None of these c
 mgvavb : GLv‡b, Nb‡Ki evûi ˆ`N©¨ a = 8 ft
 Nb‡Ki mgMÖZ‡ji †ÿÎdj = 6a2
= 6  (8)2
= 384 ft2
16 ft2
iO Ki‡Z LiP nq 36.50 UvKv
 384 ft2
,, ,, ,, ,,
36.5  384
16
UvKv
= 876 UvKv
62. If the volume of a cube is 729 cm3
, then the surface
area of the cube will be (hw` GKwU Nb‡Ki AvqZb 729 cm3
nq Z‡e Nb‡Ki mgMÖZ‡ji †ÿÎdj n‡eÑ)
a 456 cm2
b 466 cm2
c 476 cm2
d 486 cm2
d
 mgvavb : Nb‡Ki evûi ˆ`N©¨ a n‡j AvqZb a3
cÖkœg‡Z, a3
= 729
 a =
3
729 = 9 cm
 Nb‡Ki mgMÖZ‡ji †ÿÎdj = 6a2
= 6  92
= 486 cm2
63. *The surface area of a cube is 150 cm2
. Its volume is
(Nb‡Ki mgMÖZ‡ji †ÿÎdj 150 cm2
Gi AvqZb n‡eÑ)
[www.examveda.com]
a 64 cm3
b 125 cm3
c 150 cm3
d 216 cm3
b
 mgvavb : Nb‡Ki evûi ˆ`N©¨ a n‡j mgMÖZ‡ji †ÿÎdj 6a2
cÖkœg‡Z, 6a2
= 150
 a2
=
150
6
= 25  a = 25 = 5 cm
 Nb‡Ki AvqZb = a3
= 53
= 125 cm3
64. The dimensions of a piece of iron in the shape of a
cuboid are 270 cm  100 cm  64 cm. If it is melted and
recast into a cube, then the surface area of the cube will
be (Nb‡ÿÎiƒcx GKwU †jvnvi UzKivi gvÎv¸‡jv nj 270 cm  100
cm  64 cm; hw` GwU Mwj‡q GKwU NbK ˆZwi Kiv nq Z‡e
Nb‡Ki mgMÖZ‡ji †ÿÎdj n‡eÑ)
a 14400 cm2
b 44200 cm3
c 57600 cm3
d 86400 cm3
d
 mgvavb : Nb‡ÿ‡Îi AvqZb = 270  100  64 cm3
Nb‡Ki evûi ˆ`N©¨ a n‡j AvqZb = a3
cÖkœg‡Z, a3
= 270  100  64
 a =
3
27  10  100  64 =
3
33
 103
 43
= 3  10  4 = 120 cm
 wb‡Y©q mgMÖZ‡ji †ÿÎdj = 6a2
= 6  (120)2
= 86400 cm2
65. *The cost of painting the whole surface area of a cube at
the rate of 13 paise per sq. cm is 343.98 Tk. Then the
volume of the cube is : (cÖwZ eM© †m.wg. 13 cqmv nv‡i GKwU
Nb‡Ki mgMÖZj iO Ki‡Z †gvU 343.98 UvKv LiP nq| Z‡e
Nb‡Ki AvqZb n‡eÑ)
a 8500 cm3
b 9000 cm3
c 9250 cm3
d 9261 cm3
d
 mgvavb : 13 cqmv LiP nq 1 cm2
G
 34398 ,, ,, ,,
34398
13
cm2
G
= 2646 cm2
A_©vr Nb‡Ki mgMÖZ‡ji †ÿÎdj 2646 cm2
Nb‡Ki evûi ˆ`N©¨ a n‡j mgMÖZ‡ji †ÿÎdj = 6a2
 6a2
= 2646
 a2
=
2646
6
= 441
 a = 441 = 21 cm
wb‡Y©q AvqZb = a3
= (21)3
= 9261 cm3
66. An aluminium sheet 27 cm long, 8 cm broad and 1 cm
thick is melted into a cube. The difference in the
surface areas of the two solids would be (A¨vjywgwbqv‡gi
GKwU cvZ 27 cm `xN©; 8 cm cÖk¯Í Ges 1 cm cyiæ| cvZwU‡K
Mwj‡q GKwU NbK ˆZwi Kiv n‡jv| `ywU e¯‘i g‡a¨ mgMÖZ‡ji
cv_©K¨ n‡eÑ)
a Nil b 284 cm2
c 286 cm2
d 296 cm2
c
 mgvavb : cvZwUi ˆ`N©¨ a = 27 cm, cÖ¯’ b = 8 cm
cyiæZ¡ c = 1 cm
myZivs cvZwUi mgMÖZ‡ji †ÿÎdj, A1 = 2(ab + bc + ca)
= 2(27  8 + 8  1 + 1  27) cm2
= 502 cm2
cvZwUi AvqZb = abc = 27  8  1 = 216 cm3
cvZwU‡K Mwj‡q 1 evûwewkó NbK ˆZwi Ki‡j
NbKwUi AvqZb, l3
= 216
ev, l =
3
216 = cm
 NbKwUi mgMÖZ‡ji †ÿÎdj, A2 = 6l2
= 6  62
= 216 cm2
 mgMÖZ‡ji cv_©K¨, A1 – A2 = (502 – 216) = 286 cm2
67. The length of an edge of a hollow cube open at one face
is 3 metres. What is the length of the largest pole that
it can accommodate? (GKwU GK gyL †Lvjv Nb‡Ki GKwU evûi
ˆ`N©¨ 3 wgUvi| NbKwU meP‡q eo KZ ˆ`‡N©¨i GKwU †cvj‡K
RvqMv w`‡Z cvi‡e?)
a 3 m b 3 m c 3 3 m d
3
3
m b
 mgvavb : me‡P‡q eo †Mvj‡K Nb‡Ki g‡a¨
ivL‡Z n‡j Gi KY© eivei ivL‡Z n‡e| Nb‡Ki
KY© eM©‡ÿ‡Îi K‡Y©i gZ bq eis wKQzUv eo|
wP‡Îi NbKwUi KY© OD wbY©q Kwi| a
a
A
O
C
D
B

OABC eM©‡ÿ‡Îi KY© OB hv Avgiv cx_v‡Mviv‡mi m~Î †_‡K wbY©q
Ki‡Z cvwi|
OB = OA2
+ AB2
= a2
+ a2
= 2a
OBD G OD AwZfzR.
 OD = OB3
+ BD3
= (2a)2
+ a2
= 3a
GLv‡b, a = 3 m
 wb‡Y©q †cv‡ji ˆ`N©¨ = 3a = 3 3 = 3 m
68. What is the volume of a cube (in cubic cm) whose
diagonal measures 4 3 cm? (4 3 KY©wewkó Nb‡Ki AvqZb
KZ?)
a 8 b 16 c 27 d 64 d
 mgvavb : Avgiv Rvwb, a evûwewkó Nb‡Ki K‡Y©i ˆ`N©¨ 3a
GLv‡b, 3a = 4 3
 a = 4 cm
wb‡Y©q AvqZb = a3
= (4)3
= 64 cm3
69. If the total length of diagonals of a cube is 12 cm, then
what is the total length of the edges of the cube? (hw`
Nb‡Ki †gvU K‡Y©i ˆ`N©¨ 12 cm nq Z‡e NbKwUi evû¸‡jvi †gvU
ˆ`N©¨ KZ?) [www.examveda.com]
a 6 3 cm b 12 cm c 15 cm d 12 3 cm d
 mgvavb : Avgiv Rvwb, GKwU Nb‡K †gvU 4wU KY© I 12wU evû Av‡Q|
K‡Y©i ˆ`N©¨ d n‡j, 4d = 12  d = 3 cm
evûi ˆ`N©¨ a n‡j 3a = d = 3  a =
3
3
= 3 cm
 †gvU evû¸‡jvi ˆ`N©¨ = 12a = 12 3 cm
70. If the surface area of a cube is 13254 cm2
, then the
length of its diagonal is (hw` Nb‡Ki mgMÖc„‡ôi †ÿÎdj
13254 eM© †m.wg. nq, Zvn‡j H Nb‡Ki K‡Y©i ˆ`N©¨ KZ?)
a 44 3 b 45 3 cm c 46 3 d 47 3 cm d
 mgvavb : awi, Nb‡Ki evûi ˆ`N©¨ = a †m.wg.
Zvn‡j, mgMÖc„‡ôi †ÿÎdj; 6a2
= 13254  a2
= 2209  a = 47
 K‡Y©i ˆ`N©¨ = 3a = 47 3 †m.wg.
71. V1, V2, V3 and V4 are the volumes of four cubes of side
lengths x cm, 2x cm, 3x cm and 4x cm respectively.
Some statements regarding these volumes are given
below :
1. V1 + V2 + 2V3 < V4
2. V1 + 4V2 + V3 < V4
3. 2(V1 + V3) + V2 = V4
Which of these statement are correct? (V1, V2, V3 Ges
V4 AvqZb wewkó Nb‡Ki evûi ˆ`N©¨ h_vµ‡g x, 2x, 3x I 4x|
AvqZb msµvšÍ wKQz kZ© wb¤œiƒc:) [www.examveda.com]
1. V1 + V2 + 2V3 < V4
2. V1 + 4V2 + V3 < V4
3. 2(V1 + V3) + V2 = V4
a 1 and 2 b 2 and 3 c 1 and 3 d 1, 2 and 3 d
 mgvavb : V1 = x3
V2 = (2x)3
= 8x3
V3 = (3x)3
= 27x3
V4 = (4x)3
= 64x3
V1 + V2 + 2V3 = x3
+ 8x3
+ 2  273
x
= x3
+ 8x3
+ 54x3
= 63x3
< 64x3
 (1) bs kZ© mwVK
Avevi, V1 + 4V2 + V3 = x3
+ 4  8x3
+ 27x3
= x3
+ 32x3
+ 27x3
= 60x3
< 64x3
 (2) bs kZ© mwVK
2(V1 + V3) + V2 = 2(x3
+ 27x3
) + 8x3
= 2  28x3
+ 8x3
= 56x3
+ 8x3
= 64x3
= V4
 (3) bs kZ© I wVK
72. From a cube of side 8m, a square hole of 3m side is
hollowed from end to end. What is the volme of the
remaining solid? (8 m c¦vk©wewkó GKwU Nb‡Ki GK cvk †_‡K
Ab¨ GK cvk ch©šÍ GKwU 3m ˆ`N©¨ wewkó eM©vK…wZi duvcv ˆZwi
Ki‡j, Nb‡Ki Aewkó AvqZb KZ?) [www.examveda.com]
a 440 m3
b 480m3
c 508 m3
d 520 m3
a
 mgvavb : Nb‡Ki AvqZb = (8)3
= 512 Nb wg.
3wg. ˆ`N©¨ wewkó eM©vK…wZi c„‡ôi †ÿÎdj = 32
eM© wg. = 9 eM©wg.
duvcvi ˆ`N©¨ = Nb‡Ki evûi ˆ`N©¨ = 8wg.
 AvqZKvi duvcvi AvqZb = 8  9 = 72 Nb wg.
 Aewkó AvqZb = (512 – 72) = 440 Nb wg. = 440 m3
73. If the numbers representing volume and surface area
of a cube are equal, then the length of the edge of the
cube in terms of the unit of measurement will be. (hw`
GKwU Nb‡Ki AvqZb I mgMÖc„‡ôi †ÿÎd‡ji gvb GKB nq,
Zvn‡j Nb‡Ki evûi ˆ`N©¨ KZ GKK?)
a 3 b 4 c 5 d 6 d
 mgvavb : awi, Nb‡Ki evûi ˆ`N©¨ = a GKK
Zvn‡j, mgMÖ c„‡ôi †ÿÎdj = AvqZb
 6a2
= a3
 a = 6 [KviY, Nb‡Ki †ÿÎ a abvZ¥K ev¯Íe]
 evûi ˆ`N©¨ 6 GKK
74. The volume of a cube is numerically equal to the sum
of its edges. What is its total surface area in square
units? (hw` GKwU Nb‡Ki AvqZb Zvi avi¸‡jvi ˆ`‡N©¨i
†hvMd‡ji mgvb gvb nq Zvn‡j Nb‡Ki mgMÖc„‡ôi †ÿÎdj KZ?)
a 36 b 66 c 72 d 183 c
 mgvavb : awi, Nb‡Ki GKwU av‡ii ˆ`N©¨ a
Zvn‡j AvqZb = a3
†gvU av‡ii ˆ`N©¨ = 12a
cÖkœg‡Z, a3
= 12a  a2
= 12
 mgMÖZ‡ji †ÿÎdj = 6a2
= 6  12 = 72
75. Except for one face of given cube, identical cubes are
glued through their faces to all the other faces of the
given cube. If each side of the given cube measures 3
cm, then what is the total surface area of the solid body
thus formed? (hw` 3 cm evû wewkó GKwU Nb‡Ki GK c„ô ev‡`
me c„‡ô GKB iKg NbK AuvVv w`‡q jvMv‡bv nq Zvn‡j H e¯‘wUi
mgMÖ c„‡ôi †ÿÎdj KZ?) [www.examveda.com]
a 225 cm2
b 234 cm2
c 270 cm2
d 279 cm2
b
 mgvavb : Nb‡Ki 5wU c„‡ô jvMv‡j 5wU Nb‡Ki c„‡ôi †ÿÎdj n‡eÑ
5  5a2
= 5  5  (3)2
= 25  9 = 225 cm2
[GLv‡b, GKwU c„ô AvVv w`‡q jvMv‡jv ZvB 6a2
Gi e`‡j 5a2
n‡e]
Avi cÖ_g Nb‡Ki Aci c„‡ôi †ÿÎdj = a2
= 9 cm3
 †gvU †ÿÎdj = 225 + 9 = 234 cm2
76. A solid cube just gets completely immersed in water
when a 0.2 kg mass is placed on it. If the mass is
removed, the cube is 2 cm above the water level. What
is length of each side of the cube? (hw` GKwU Nb‡Ki Ici
2kg fi †`qv nq Zvn‡j NbKwU cvwb‡Z m¤ú~Y© wbgw¾Z _v‡K| Avi
hw` fi mwi‡q †bqv nq Zvn‡j NbKwU 2 cm cvwbi Dc‡i _v‡K|)
[www.examveda.com]
a 6 cm b 8 cm c 10 cm d 12 cm c
 mgvavb : awi, Nb‡Ki evûi ˆ`N©¨ = a cm
Zvn‡j Nb‡Ki c„‡ôi AvqZb = a2
cm2
e¯‘i fi = NbK Øviv AcmvwiZ cvwbi fi
= NbK Øviv AcmvwiZ cvwbi AvqZb  cvwbi NbZ¡
= Nb‡Ki c„‡ôi †ÿÎdj  f‡ii Rb¨ D”PZvi cwieZ©b  cvwbi NbZ¡
[e¯‘i fi = 0.2kg = 0.2  1000 g = 200 g]
 200 = a2
 2  1 [cvwbi NbZ¡ =
1g
cc
]
 2a2
= 200
 a2
= 100  a = 10

77. A cube of length 1 cm is taken out from a cube of
length 8 cm. What is the weight of the remaining
portion? (8 cm evûwewkó NbK †_‡K 1 cm evûwewkó NbK †ei
K‡i wb‡j Aewkó As‡ki IRb KZ?) [www.examveda.com]
a
7
8
of the weight of the original cube
b
8
9
c
63
64
d
511
512
of the weight of the original cube d
 mgvavb : awi, Nb‡Ki NbZ¡ = P g/cc
eo Nb‡Ki AvqZb = 83
= 512 cm3
†QvU Nb‡Ki AvqZb = 13
= 1 cm3
Aewkó As‡ki AvqZb = (512 – 1) = 511 cm3
eo Nb‡Ki fi = 512P
Aewkó As‡ki fi = 511P
 Aewkó Ask g~j As‡ki,
511p
512p
=
511
512
Ask
78. *How many cubes of 10 cm edge can be put in a cubical
box of 1 m edge? (10 cm avi wewkó KZ¸‡jv NbK 1 m
aviwewkó NbvK…wZi e‡· ivLv hv‡e?) [www.examveda.com]
a 10 b 100 c 1000 d 10000 c
 mgvavb : 10 cm avi wewkó Nb‡Ki AvqZb = (10)3
= 1000 cm3
1 m ev 100 cm avi wewkó Nb‡Ki AvqZb = (100)3
cm3
= 1000000 cm3
 NbK ivLv hv‡e
1000000
1000
= 1000wU
79. A 4 cm cube is cut into 1 cm cubes. The total surface
area of all the small cubes is (4 cm aviwewkó NbK‡K 1 cm
aviwewkó wKQz Nb‡K cwiYZ Kiv n‡j, †QvU NbK¸‡jv †gvU c„‡ôi
†ÿÎdj KZ?) [www.examveda.com]
a 24 cm2
b 96 cm2
c 384 cm2
d None of these c
 mgvavb : 4 cm avi wewkó Nb‡Ki AvqZb = 43
= 64 cm3
1 cm avi wewkó Nb‡Ki AvqZb = 13
= 1 cm3
 †gvU
64
1
ev, 64wU 1 cm3
AvqZb wewkó NbK n‡e|
Nb‡Ki c„‡ôi †ÿÎdj = 6a2
 1 cm avi wewkó 1wU Nb‡Ki c„‡ôi †ÿÎdj = 6 cm2
1 cm avi wewkó 64wU Nb‡Ki c„‡ôi †ÿÎdj = 64  6 cm2
= 384 cm2
80. A rectangular block with a volume of 250 cm3
was
sliced into two cubes of equal volume. How much
greater (in sq. cm) is the combined surface area of the
two cubes then the original surface area of the
rectangular block? (GKwU AvqZvK…wZi Nbe¯‘‡K hvi AvqZb
250 cm3
†K‡U `yBwU GKB Nb‡K cwiYZ Kiv nj| Zvn‡j NbK
`yBwUi wgwjZ †ÿÎdj AvqZKvi Nbe¯‘i †P‡q KZ †ekx n‡e?)
a 48.64 b 50 c 56.25 d 84.67 b
 mgvavb : †h‡nZz AvqZKvi e¯‘‡K †K‡U Nb‡K cwiYZ Kiv nq ZvB
Nb‡Ki avi, a n‡e AvqZKvi e¯‘i D”PZv, h Gi A‡a©K|
†h‡nZz AvqZKvi e¯‘ KvUvi Rb¨ NbK nq ZvB Gi Aci evû `yBwU
GKB n‡e Ges Zv a|
 AvqZKvi Nbe¯‘i D”PZv, h = 2a
 AvqZb = a  a  h
 250 = a  a  2a  250 = 2a3
a3
= 125  a = 5
 AvqZKvi Ne¯‘i c„‡ôi †ÿÎdj = 2(a  h + a  h + a  a)
= 2{a  2a + a  2a + a  a}
= 2{2a2
+ 2a2
+ a2
}
= 2  5a2
= 10a2
= 10  (5)2
= 10  25 = 250 cm2
GKwU Nb‡Ki c„‡ôi †ÿÎdj = 6a2
= 6  (5)2
= 6  25 = 150
 `yBwU Nb‡Ki c„‡ôi †gvU †ÿÎdj = 2  150 = 300 cm2
 †ÿÎd‡ji cwieZ©b = (300 – 250) = 50 cm2
81. A rectangular box measures internally 1.6 m long. 1 m
broad and 60 cm deep. The number of cubical blocks
each of edge 20 cm that can be packed inside the box is
(GKwU AvqZKvi ev‡·i ˆ`N©¨ 1.6 m cÖ¯’ 1 m Ges D”PZv 60 cm.
n‡j 20 cm aviwewkó KZ¸‡jv NbK G‡Z ¯’vb cv‡e?)
a 30 b 53 c 60 d 120 d
 mgvavb :
AvqZvKvi ev‡·i AvqZb = ˆ`N©¨  cÖ¯’  D”PZv = abc
= 1.6  1  0.6 m3
a = 1.6 m
= 0.96 m3
b = 1 m
c = 60 cm = 0.6m
Nb‡Ki evûi ˆ`N©¨ = 20 cm = 0.2 m
 Nb‡Ki AvqZb = (0.2)3
m3
= 0.008 m3
 ev‡·
0.96
0.008
ev, 120wU NbK‡K ¯’vb †`qv hv‡e|
82. How many cubes of 3 cm edge can be cut out of a cube
of 18 cm edge? (18 cm aviwewkó NbK †_‡K 3 cm avi wewkó
KZ¸‡jv NbK KvUv hv‡e?)
a 36 b 216 c 218 d 432 b
 mgvavb : 18 cm avi wewkó Nb‡Ki AvqZb = (18)3
cm3
3 cm avi wewkó Nb‡Ki AvqZb = (3)3
cm3
 KvUv hv‡e †gvU
(18)3
(3)3 ev, (6)3
ev, 216wU NbK
83. Shobhraj takes a cube of 1 m edge-length and
meticulously cuts smaller cubes, each of edge-length 1
mm from the parent cube. He joins these small cubes
end-to-end. Thus, the total length of this cube-robe will
be (me&&ivR 1 m avi wewkó NbK‡K †K‡U 1 mm avi wewkó Nb‡K
cwiYZ K‡i| hw` GB †QvU †QvU NbK¸‡jv GKwUi ci GKwU emv‡bv
nq, Zvn‡j Gi ˆ`N©¨ KZ n‡e?)
a 1 km b 10 km c 100 km d 1000 km d
 mgvavb : 1 m avi wewkó Nb‡Ki AvqZb = 13
m3
1 mm ev,
1
1000
ev 0.001 m avi wewkó
Nb‡Ki AvqZb = (0.001)3
m3
 †QvU NbK n‡e,
1
(0.001)3 ev 109
wU
cÖwZwU Nb‡Ki ˆ`N©¨ = 1 mm = 0.001 m
 †gvU Nb‡Ki ˆ`N©¨ = 109
 0.001 m
= 106
m =
106
1000
km = 1000 km
84. How many small cubes, each of 96 cm surface area, can
be formed from the material obtained by melting a
larger cube of 384 cm surface area? (384 eM© †m.wg.
†ÿÎd‡ji †Kvb NbK‡K Mwj‡q 96 eM© †m.wg. †ÿÎdj wewkó KZwU
NbK MVb Kiv m¤¢e?)
a 5 b 8 c 800 d 8000 b
 mgvavb : g‡b Kwi, †QvU Nb‡Ki avi a Ges eo Nb‡Ki avi b
Avgiv Rvwb,
Nb‡Ki mgMÖZ‡ji †ÿÎdj = 6  (Nb‡Ki avi)2
cÖkœg‡Z, 6a2
= 96
 a2
= 16
 a = 4
Ges 6b2
= 384
 b2
= 64
 b = 8

 †QvU Nb‡Ki AvqZb = 43
= 64
Ges eo Ó Ó = 83
= 512
 NbK evbv‡bv hv‡e =
512
64
= 8wU
85. *The volume of a cuboid is twice that of a cube. If the
dimensions of the cuboid are 9 cm, 8 cm and 6 cm, the
total surface area of the cube is (GKwU Nbe¯‘i AvqZb Aci
GKwU Nb‡Ki AvqZ‡bi wØ¸Y| hw` Nbe¯‘i gvÎv¸‡jv h_vµ‡g, 9
cm, 8 cm I 6 cm nq, Zvn‡j Nb‡Ki mgMÖZ‡ji †ÿÎdj KZ?)
[www.examveda.com]
a 72 cm2
b 108 cm2
c 216 cm2
d 432 cm2
c
 mgvavb : Nbe¯‘i AvqZb = abc
= 9  8  6 = 432 cm3
Avevi, Nb‡Ki AvqZb = (evûi ˆ`N©¨)3
cÖkœg‡Z,
Nb‡Ki AvqZb Nbe¯‘i AvqZb Gi A‡a©K 
432
2
 216 cm3
 (evûi ˆ`N©¨)3
= 216
 evûi ˆ`N©¨ = 6
 Nb‡Ki mgMÖc„‡ôi †ÿÎdj = 6  (evûi ˆ`N©¨)2
= 6  (6)2
= 216 cm2
86. A cuboidal block of 6 cm  9 cm  12 cm is cut up into an
exact number of equal cubes. The least possible
number of cubes will be (GKUv 6m  9 cm  12 cm m¤ú~Y©
Nbe¯‘‡K †K‡U c~Y©msL¨vq Nb‡K cwiYZ Kiv n‡j me©wb¤œ KZ¸‡jv
NbK cvIqv †h‡Z cv‡i?)
a 6 b 9 c 24 d 30 c
 mgvavb : †h‡nZz me©wb¤œ msL¨K NbK PvIqv n‡q‡Q ZvB m‡e©v”P
AvqZ‡bi NbK KvU‡Z n‡e hv‡Z AvqZKvi e¯‘i m¤ú~Y© eëüZ nq|
Nbe¯‘i, avi¸‡jv a = 6 cm
b = 9 cm
c = 12 cm
 Nb‡Ki avi n‡e, a, b I c Gi M.mv.¸. Gi gv‡bi mgvb|
myZivs 6, 9 I 12 Gi M.mv.¸. = 3
Zvn‡j Gi AvqZb = (3)3
Nb †m.wg. = 27 Nb †m.wg.
Avevi Nbe¯‘i AvqZb = 6  9  12 Nb †m.wg. = 648 Nb †m.wg.
 me©wb¤œ KvUv hv‡e
648
27
 24wU NbK
87. The size of a wooden block is 5  10  20 cm. How
many such blocks will be required to construct a solid
wooden cube of minimum size? (hw` Kv‡Vi eø‡Ki AvKvi 5
 10  20 cm nq Zvn‡j me©wb¤œ KZwU eøK eënvi K‡i GKwU NbK
ˆZwi Kiv hv‡e?)
a 6 b 8 c 12 d 16 b
 mgvavb : Kv‡Vi eø‡Ki avi¸‡jv h_vµ‡g 5, 10 I 20 Ges G‡ì
j.mv.¸. = 20
Zvn‡j, †h NbK n‡e Zvi avi n‡e 20 cm
Gi AvqZb = (20)3
= 8000 cm3
Avevi Kv‡Vi eø‡Ki AvqZb = 5  10  20 = 1000 cm3
 Kv‡Vi UzKiv jvM‡e =
8000
1000
 8wU
88. An iron cube of side 10 cm is hammered into a
rectangular sheet of thickness 0.5 cm. If the side of the
sheet are in the ratio 1 : 5, thd sides are (10 †mwg
aviwewkó GKwU †jvnvi NbK‡K wcwU‡q 0.5 †mwg cyiæ GKwU
AvqZKvi Pvì evbv‡bv nj| hw` PvìwUi ˆ`N©¨ I cÖ‡¯’i AbycvZ 1
: 5 nq, Z‡e Zvi avi¸wj njÑ)
a 10 cm, 50 cm b 20 cm, 100 cm
c 40 cm, 200 cm d None of these b
 mgvavb : †jvnvi Nb‡Ki avi, a = 10 cm
 AvqZb = (10)3
 1000 cm3
†jvnvi NbK‡K wcwU‡q AvqZKvi cv‡Z cwiYZ Kiv nq
hvi cyiæZ¡ c = 0.5 cm
Aci gvb¸‡jv AbycvZ 1 : 5
A_©vr GKwUi gvb, a = x
Ges AciwUi gvb, b = 5x
 AvqZb = x  5x  0.5 = 1000
 x2
= 400  x = 20
 Aci gvÎv¸‡jv h_vµ‡g 20 cm I 20  5  100 cm
89. A cube of white chalk is painted red, and then cut parallel
to the sides to form two rectangular solids of equal
volumes. What percent of the surface area of each of
the new solids is not painted red? (GKwU NbvKvi mv`v PK‡K
jvj iO Kiv nj| G‡K cÖ‡¯’i mgvšÍiv‡j †K‡U `ywU GKB AvqZ‡bi
AvqZvKvi `ªe¨ evbv‡bv nj| cÖwZwU UzK‡ivi KZ kZvsk ivOv‡bv bq?)
a 15% b 16
2
3
% c 20% d 25% d
 mgvavb : awi, Nb‡Ki evûi ˆ`N©¨ = x
c„‡ôi †ÿÎdj = 6x2
 6x2
†ÿÎd‡ji Dci jvj iO Kiv nq|
G‡K `yB UzKiv Kivq jvj †ÿÎdj `yBfv‡M fvM n‡q hvq|
`yBwU AvqZKvi e¯‘i cÖ‡Z¨‡Ki jvj iO Kiv †ÿÎdj =
6x2
2
 3x2
cÖwZwU AvqZKvi Nbe¯‘i GKwU bZzb c„ô †hvM n‡q‡Q hv iO Kiv bq
Gi †ÿÎdj Nb‡Ki †h‡Kvb c„‡ôi †ÿÎd‡ji mgvb A_©vr x2
 AvqZKvi e¯‘i †ÿÎdj 3x2
+ x2
= 4x2
Ges iO Qvov As‡ki †ÿÎdj = x2
 jvj iO Qvov Ask =
x2
4x2 =
1
4
 25%
90. There is a cube of volume 216 cm3
. It is to be moulded
into a cuboid having one edge equal to 6 cm. The
number of ways that it can be done so that the edge
have different integral values is (GKwU Nb‡Ki AvqZb 216
Nb‡mwg, GwU †_‡K 6 †mwg GKwU aviwewkó GKwU AvqZb ˆZwi Kiv
nj| Giƒc KwU c×wZ‡Z AvqZbNbwU ˆZwi Kiv †h‡Z cv‡i hv‡Z
GUvi avi¸wji gvc wfbœ nq?)
a 1 b 2 c 3 d 4 d
 mgvavb : †Ìqv Av‡Q, Nb‡Ki AvqZb = 216 cm3
G‡K GKwU Nbe¯‘‡Z cwiYZ Ki‡Z n‡e hvi GKwU avi 6 cm
awi Aci avi `yBwU h_vµ‡g a, b
Zvn‡j, ab  6 = 216 36 = 4  9
 ab = 36 = 12  3
= 18  2
= 36  1
GLb a Ges b mgvb n‡Z cvi‡e bv AvKvi a I b c~Y©msL¨v n‡Z
n‡e, GiKg 4wU Dcvq Av‡Q|
91. If three cubes of copper, each with an edge of 6 cm 8
cm and 10 cm respectively are melted to form a single
cube, then the diagonal of the new cube will be (wZbwU
Zvgvi NbK, hv‡ì avi 6 †mwg, 8 †mwg I 10 †mwg, Mwj‡q GKwU
bZzb NbK ˆZwi Kiv n‡j, Zvi KY©‰`N©¨ KZ n‡e?)
a 18 cm b 19 cm c 19.5 cm d 20.8 cm d
 mgvavb : awi, bZzb Nb‡Ki avi = x cm
 AvqZb = x3
cm
1g Nb‡Ki avi = 6 cm
 AvqZb = (6)3
cm3
[AvqZb = (evûi ˆ`N©¨)3
]
2q Nb‡Ki avi = 8 cm
AvqZb = (8)3
cm3
AvqZb = (10)3
cm3

cÖkœg‡Z,
bZzb Nb‡Ki AvqZb = (1g NbK + 2q NbK + 3q NbK) Gi AvqZb
 x3
= 63
+ 83
+ 103
 x3
= 1728  x = 12
 Nb‡Ki K‡Y©i ˆ`N©¨ = 3x [Nb‡Ki KY© = 3a]
= 12 3 = 20.8 cm
92. *A larger cube is formed from the material obtained by
melting three smaller cubes of 3, 4, and 5 cm side. The
ratio of the total surface areas of the smaller cubes and
the larger cube is (3 †mwg, 4 †mwg I 5 †mwg aviwewkó wZbwU
†QvU NbK Mwj‡q GKwU eo NbK †Zwi Kiv n‡j, Zv‡ì mgMÖZ‡ji
†ÿÎdj I bZzb Nb‡Ki mgMÖ Z‡ji †ÿÎd‡ji AbycvZ KZ n‡e?)
a 2 : 1 b 3 : 2 c 25 : 18 d 27 : 20 c
 mgvavb : awi, eo Nb‡Ki avi = x cm
 AvqZb = x3
cm3
[Nb‡Ki c„‡ôi †ÿÎdj = 6a2
]
1g Nb‡Ki avi = 3 cm
Ges c„‡ôi †ÿÎdj = 6  (3)2
= 6  9 = 54 cm2
 AvqZb = (3)3
cm3
2q Nb‡Ki avi = 4 cm Ges c„‡ôi †ÿÎdj = 6  42
cm2
= 6  16 ev, 96 cm2
 AvqZb = (4)3
cm3
Ges c„‡ôi †ÿÎdj = 6  52
cm2
= 5  25 cm2
= 150 cm2
 AvqZb = (5)3
cm3
cÖkœg‡Z, eo Nb‡Ki AvqZb = (1g + 2q + 3q) Nb‡Ki AvqZb
 x3
= 33
+ 43
+ 53
 x3
= 27 + 64 + 125  x3
= 216
 x = 6
 Nb‡Ki c„‡ôi †ÿÎdj = 6  62
cm2
[ mgMÖ c„‡ôi †ÿÎdj = 6a2
]
= 216 cm2
†QvU NbK¸‡jvi c„‡ôi †ÿÎd‡ji †hvMdj = (54 + 96 + 150) cm2
= 300 cm2
 wb‡Y©q AbycvZ =
†QvU NbK¸‡jvi c„‡ôi †ÿÎd‡ji †hvMdj
eo Nb‡Ki c„‡ôi †ÿÎdj
=
300
216
=
100
72
=
50
36
=
25
18
= 25 : 18
93. Five equal cubes, each of side 5 cm, are placed adjacent
to each other. The volume of the new solid formed will
be (5 †mwg aviwewkó cuvPwU mgvbNbK cvkvcvwk emv‡bv nj| bZzb
e¯‘wUi AvqZb wbY©q Ki|)
a 125 cm3
b 625 cm3
c 15525 cm3
d None of these b
 mgvavb : 5wU GKB avi wewkó Nb‡Ki gv‡S †h‡Kvb GKwU Nb‡Ki
avi = 5 cm
 H Nb‡Ki AvqZb = (5)3
cm3
[Nb‡Ki AvqZb = a3
]
= 125 cm3
 5wU Nb‡Ki AvqZb = 5  125 cm3
= 625 cm3
94. *If three equal cubes are placed adjacently in a row,
then the ratio of the total surface area of the new
cuboid to the sum of the surface areas of the three
cubes will be (wZbwU mgvb NbK‡K cvkvcvwk emv‡bv n‡j, bZzb
MwVZ AvqZbNbwUi mgMÖZ‡ji †ÿÎdj I NbK¸wji mgMÖZ‡ji
†ÿÎd‡ji mgwói AbycvZ KZ n‡e?) [www.examveda.com]
a 1 : 3 b 2 : 3 c 5 : 9 d 7 : 9 d
 mgvavb : awi, cÖwZwU Nb‡Ki evûi ˆ`N©¨ = a
cÖwZwU Nb‡Ki c„‡ôi †ÿÎdj = 6a2
 3wU Nb‡Ki c„‡ôi †ÿÎdj = 3  6a2
= 18a2
3wU NbK‡K GKwUi ci GKwU mvRv‡bvi d‡j Drcbœ
Nbe¯‘i ˆ`N©¨ = 3  a Ges cÖ¯’ = a Ges D”PZv = a
Zvn‡j, Nbe¯‘i c„‡ôi †ÿÎdj n‡e, 2(a  3a + a  3a + a  a)
= 2(3a2
+ 3a2
+ a2
)
[ Nbe¯‘i c„‡ôi †ÿÎdj = 2(ab + bc + ca)]
= 2  7a2
= 14a2
 Nbe¯‘ I Nb‡Ki c„‡ôi †ÿÎd‡ji †hvMd‡ji AbycvZ n‡e,
14a2
18a2 =
7
9
= 7 : 9
95. Three cubes with sides in the ratio 3 : 4 : 5 are melted
to form a single cube whose diagonal is 12 3 cm. The
sides of the cubes are (wZbwU Nb‡Ki AbycvZ 3 : 4 : 5|
G¸wj‡K Mwj‡q bZzb GKwU NbK †Zwi Kiv n‡j, hvi KY© ˆ`N©¨
12 3 †mwg| NbK¸wji ˆ`N©¨ KZ?)
a 3 cm, 4 cm, 5 cm b 6 cm, 8 cm, 10 cm
c 9 cm, 12 cm, 15 cm d None of these b
 mgvavb : †Ìqv Av‡Q, Nb‡Ki K‡Y©i ˆ`N©¨ = 12 3 cm
awi, Nb‡Ki evûi ˆ`N©¨ = a
Zvn‡j KY© = 3a
 3a = 12 3
 a = 12
 Nb‡Ki evûi ˆ`N©¨ = 12 cm;  AvqZb = (12)3
cm3
wZbwU Nb‡Ki av‡ii AbycvZ = 3 : 4 : 5
Zvn‡j, cÖ_g Nb‡Ki evûi ˆ`N©¨ = 3x
2q Nb‡Ki avi = 4x
3q ,, ,, = 5x
 cÖ_g Nb‡Ki AvqZb = (3x)3
[Nb‡Ki AvqZb = a3
]
2q Nb‡Ki AvqZb = (4x)3
3q Nb‡Ki AvqZb = (5x)3
†QvU NbK¸‡jvi AvqZ‡bi †hvMdj eo Nb‡Ki AvqZ‡bi mgvb n‡e|
A_©vr (3x)3
+ (4x)3
+ (5x)3
= (12)3
 27x3
+ 64x3
+ 125x3
= 1728  216x3
= 1728
 x3
= 8  x = 2
 1g wUi avi = 3x = 3  2 = 6 cm
2q wUi avi = 4x = 4  2 = 8 cm
3q wUi avi = 5x = 5  2 = 10 cm
96. If the volumes of two cubes are in the ratio 27 : 1, the
ratio of their edges is (`ywU Nb‡Ki AvqZ‡bi AbycvZ 27 : 1,
Zvi ˆ`‡N©ï AbycvZ KZ?)
a 1 : 3 b 1 : 27 c 3 : 1 d 27 : 1 c
 mgvavb : awi, cÖ_g Nb‡Ki evûi ˆ`N©¨ = a GKK
Zvn‡j, AvqZb = a3
Nb GKK
Avevi awi, 2q Nb‡Ki evûi ˆ`N©¨ = b GKK
 AvqZb = b3
Nb GKK
 AvqZ‡bi AbycvZ = a3
: b3
hv cÖkœg‡Z 27 : 1

a3
b3 =
27
1




a
b
3
=
33
1
[
am
bm =



a
b
m
]

a
b
=
3
1
 a : b = 3 : 1
 av‡ii AbycvZ 3 : 1
97. The volume of two cubes are in the ratio 8 : 27. The
ratio of their surface areas is (`ywU AvqZ‡bi AbycvZ 8 : 27|
Zv‡ì mgMÖZ‡ji †ÿÎd‡ji AbycvZ?) [www.examveda.com]
a 2 : 3 b 4 : 9
c 12 : 9 d None of these b

 mgvavb : awi, GKwU Nb‡Ki avi = a
 AvqZb = a3
Ges mgMÖ c„ôZ‡ji †ÿÎdj = 6a2
Avevi awi,
Aci Nb‡Ki avi = b
 AvqZb = b3
Ges mgMÖ c„ôZ‡ji †ÿÎdj = 6b2
NbK `yBwUi AvqZ‡bi AbycvZ = a3
: b3
hv cÖkœg‡Z 8 : 27
A_©vr a3
: b3
= 8 : 27

a3
b3 =
8
27




a
b
3
=



2
3
3

a
b
=
2
3
... ... (i)
Avevi, mgMÖc„‡ôi †ÿÎd‡ji AbycvZ = 6a2
: 6b2
=
6a2
6b2 =
a2
b2
=



2
3
2
[(i) bs n‡Z gvb ewm‡q]
=
4
9
= 4 : 9
98. Two cubes have volumes in the ratio 1 : 27. Then the
ratio of the area of the face of one of the cubes to that
of the other is (`ywU N‡bi AvqZ‡bi AbycvZ 1 : 27, Z‡e
Nb`ywUi GK cv‡k¦©i †ÿÎd‡ji AbycvZ KZ?)
a 1 : 3 b 1 : 6 c 1 : 9 d 1 : 12 c
 mgvavb : awi, GKwU Nb‡Ki avi = a
 AvqZb = a3
†h‡Kvb GK c„‡ôi †ÿÎdj = a2
Avevi awi, 2q Nb‡Ki avi = b
 AvqZb = b3
†h‡Kvb GK c„‡ôi †ÿÎdj = b2
 AvqZ‡bi AbycvZ = a3
: b3
cÖkœg‡Z hv 1 : 27
A_©vr
a3
b3 =
1
27

a
b
=
1
3
 NbK `yBwU †h‡Kvb GK c„‡ôi †ÿÎd‡ji AbycvZ =
a2
b2 =



a
b
2
=



1
3
2
=
1
9
= 1 : 9
99. If each edge of a cube is doubled, then its volume (hw`
†Kvb N‡bi cÖwZwU cvk‡K wØ¸Y Kiv nq, Z‡e AvqZb n‡e)
a is doubled b becomes 4 times
c becomes 6 times d becomes 8 times d
 mgvavb : awi, avi = a GKK
 AvqZb = a3
Nb GKK
avi 2 ¸Y Ki‡j avi = 2  a GKK
= 2a GKK
ZLb AvqZb = (2a)3
Nb GKK
= 8a3
Nb GKK
 AvqZb nq c~‡e©i
8a3
a3  8 ¸Y
100. By what percent the volume of a cube increases if the
length of each edge was increased by 50%? (hw` †Kvb
Nb‡Ki cÖwZwU cvk‡K 50% e„w× Kiv nq, Z‡e Gi AvqZb KZ
kZvsk e„w× cv‡e?)
a 25% b 125% c 237.5% d 273.5% c
 mgvavb : awi, Nb‡Ki avi = a GKK
AvqZb = a3
Nb GKK
50% e„w×‡Z avi =



a + a Gi
50
100
GKK
= a +
a
2
GKK =
3a
2
GKK
 AvqZb =



3a
8
3
=
27
4
a3
Nb GKK
 AvqZb e„w× =



27a3
8
– a3
=
27a3
– 8a3
8
=
19a3
8
 a3
Nb GKK †_‡K AvqZb e„w× cvq
10
8
a3
Nb GKK
 1 Nb GKK ,, ,, ,, ,,
19 a3
8 a3 Nb GKK
 100 Nb GKK ,, ,, ,, ,,
19a3
 100
8a3 Nb GKK
= 237.5%
101. If each edge of a cube is increased by 25%, then the
percentage increase in its surface area is : (hw` GKwU
Nb‡Ki cÖwZwU evûi ˆ`N©¨ 25% e„w× cvq Z‡e Gi mgMÖZ‡ji
†ÿÎdj KZ kZKiv e„w× cv‡e?)
a 25% b 48.75% c 50% d 56.25% d
 mgvavb : Nb‡Ki evûi ˆ`N©¨ a n‡j mgMÖZ‡ji †ÿÎdj A = 6a2
evûi ˆ`N©¨ 25% e„w× †c‡j bZzb ˆ`N©¨ l = a + 25% a
= a +
25
100
a =
5
4
a
 bZzb mgMÖZ‡ji †ÿÎdj A2 = 6l2
= 6 



5
4
a
2
=
75
8
a2
†ÿÎdj e„w× A2 – A1 =
75
8
a2
– 6a2
=
27
8
a2
6a2
G‡Z e„w× cvq
27
8
a2
 1 ,, ,, ,,
27
8
a2

1
6a2
 100 ,, ,, ,,
27 a2
 100
8  6a2 = 56.25
 wb‡Y©q e„w× 56.25%
102. A cube of edge 20 cm is completely immersed in a
rectangular vessel containing water. If the dimensions
of the base of the vessel are 20 cm by 40 cm, the rise in
water level will be (GKwU 20 cm evûwewkó NbK‡K AvqZvKvi
GKwU cvwbi cv‡Î Wzev‡bv nj| hw` cv‡Îi wb‡Pi Z‡ji cwigvc 20
cm I 40 cm nq Z‡e, cvwb KZ D”PZvq DV‡e?)
 mgvavb : GLv‡b, NbKwU‡K cvwb‡Z m¤ú~Y©iƒ‡c Wzev‡j AcmvwiZ
cvwbi AvqZb NbKwUi AvqZ‡bi mgvb n‡e|
NbKwUi AvqZb = (20)3
cm3
= 8000 cm3
cvwbi D”PZv h n‡j,
h  20  40 = 8000
 h =
8000
20  40
= 10 cm
103. A circular well with a diameter of 2 meters, is dug to a
depth of 14 metres. What is the volume of the earth
dug out? (2m e¨vm I 14 m MfxiZvm¤úbœ GKwU mge„Ë f‚wgK
K‚c Lbb Kiv n‡jv| LbKK…Z gvwUi AvqZb KZ?) [www.examveda.com]
a 32 m3
b 36 m3
c 40 m3
d 44 m3
d
 mgvavb : mge„Ëf‚wgK K‚c wmwjÛvivK…wZi hviÑ
e¨vm, 2r = 2 m
 r = 1 m
D”PZv, h = 14 m
 LbbK…Z gvwUi AvqZb = r2
h
=
22
7
 12
 14 =
22
7
 14 m3
= 44 m3

104. Find the cost of a cylinder of radius 14 m and height
3.5 m when the cost of its metal is 50 Tk. per cubic
metre. (14 m e¨vmva© I 3.5 m D”PZvm¤úbœ GKwU avZe wmwjÛvi
wbg©v‡Y LiP KZ hw` cÖwZ NbwgUv‡i H avZzi g~j¨ 50 UvKv nq?)
[www.examveda.com]
a 100208 Tk. b 107800 Tk.
c 10800 Tk. d 109800 Tk. b
 mgvavb : e¨vmva©, r = 14 m
D”PZv, h = 3.5 m
AvqZb = r2
h
=
22
7
 142
 3.5 m3
= 2156 m3
†gvU LiP = 2156  50 UvKv
= 107800 UvKv
105. If the radius and height of a right circular cylinder are
21 cm and 35 cm respectively, then the total surface
area of the cylinder is (GKwU mge„Ëf‚wgK wmwjÛv‡ii e¨vmva© I
D”PZv h_vµ‡g 21 cm I 35 cm n‡j Gi mgMÖ c„‡ôi †ÿÎdjÑ)
a 7092 sq cm b 7192 sq cm
c 7292 sq cm d 7392 sq cm d
 mgvavb : e¨vmva©, r = 21 cm
D”PZv, h = 35 cm
mgMÖ c„‡ôi †ÿÎdj = 2r (r + h)
= 2 
22
7
 21  (21 + 35) cm2
= 7392 cm2
106. The capacity ofa cylindrical tank is 246.4 litres. If the
hight is 4 metres, what is the diameter of the base?
(wmwjÛvivK…wZi GKwU U¨vsK Gi aviYÿgZv 246.4 wjUvi| D”PZv
4 m n‡j f‚wgi e¨vm e¨vm KZ?) [www.examveda.com]
a 1.4 m b 2.8 m c 14 m d 28 m
e None of these
 mgvavb : D”PZv, h = 4 m
e¨vmva© r n‡j
aviYÿgZv ev AvqZb = r2
h
cÖkœg‡Z, r2
h = 264.4 
1
1000
; [1000 litres = 1 m3
]

22
7
 r2
 4 =
264.4
1000
 r2
=
7  264.4
1000  22  4
 r =
7  264.4
1000  22  4
 r = 0.145 m  r = 14.5 cm
 e¨vm = 2  14.5 cm = 29 cm
107. The volume of a right circular cylinder, 14 cm in the
height is equal to that of a cube whose edge is 11 cm.
The radius of the base of the cylinder is (14 cm
D”PZvwewkó GKwU mge„Ëf‚wgK wmwjÛv‡ii AvqZb 11 cm aviwewkó
GKwU Nb‡Ki AvqZ‡bi mgvb| Gi f‚wgi e¨vmva©Ñ)
[www.examveda.com]
a 5.2 cm b 5.5 cm c 11 cm d 22 cm b
 mgvavb : awi, e¨vmva© = r
D”PZv, h = 14 cm
 AvqZb = r2
h
x = 11 cm aviwewkó Nb‡Ki AvqZb = 113
cm3
cÖkœg‡Z, r2
h = 113

22
7
 r2
 14 = 113
 44r2
= 113
 r2
=
113
4  11
 r2
=
121
4
 r =
121
4
 r =
11
2
 r = 5.5 cm
108. Capacity of a cylindrical vessel is 25.872 litres. If the
height of the cylinder is three times the radius of its
base, what is the area of the base? (wmwjÛvivK…wZi GKwU
cv‡Îi aviYÿgZv 25.872 wjUvi| D”PZv hw` f‚wgi e¨vmv‡a©i
wZb¸Y nq, Z‡e f‚wgi †ÿÎdj KZ?) [www.examveda.com]
a 336 cm2
b 616 cm2
c 1232 cm2
d Cannot be determined
e None of these b
 D”PZv, h = 3r
AvqZb/aviY ÿgZv = r2
h = r2
(3r) = 3r3
cÖkœg‡Z, 3r3
= 25.872 
1
1000
[ 1000 wjUvi = 1 m3
]
 r3
=
25.872
3

1
1000
 r3
=
25.872
3 
22
7
 1000
 r3
=
343
125  1000
 r3
=



7
50
3
 r =
7
50
 f‚wgi †ÿÎdj = r2
=
22
7




7
50
2
=
22  7
2500
m2
= 0.0616 cm2
= 0.0616  10000 cm2
[ 1 m2
= 10000 cm2
]
= 616 cm2
109. *Two rectangular sheets of paper, each 30 cm  18 cm
are made into two right circular cylinders, one by
rolling the paper along its length and the other along
the breadth. The ratio of the volumes of the two
cylinders, thus formed is (30 cm  18 cm gv‡ci `ywU
AvqZvKvi KvM‡Ri kx‡Ui GKwU‡K ˆ`N©¨ eivei I Ab¨wU‡K cÖ¯’ eivei
gywo‡q `ywU mge„Ëf‚wgK wmwjÛv‡i cwiYZ Kiv n‡jv| wmwjÛviØ‡qi
AvqZ‡bi AbycvZÑ) [www.examveda.com]
a 2 : 1 b 3 : 2 c 4 : 3 d 5 : 3 d
 mgvavb : ˆ`N©¨ eivei,
e¨vmva©, r1 n‡j, 2r1 = 30
r1 =
30
2
18cm
30 cm
 h1=18cm
cwiwa=30cm
 AvqZb, V1 = r1
2
h1
=  



30
2
2
 18 cm2
=
152

 18 cm2
GKBfv‡e, wØZxq wmwjÛv‡ii e¨vmva© r2 n‡j
2r2 = 18  r2 =
9

h2 = 30 cm
AvqZb, V2 = r2
2
h2
=  



9

2
 30 cm2
=
92
 30

V1
V2
=
(152
 18)

(92
 30)

=



15
9
2

18
30
=



5
3
2

18
30
=
25
9

18
30
=
5
3
110. Three rectangle A1, A2 and A3 have the same area.
Their lengths a1, a2 and a3 respectively are such that a1
< a2 < a3. Cylinders C1, C2 and C3 are formed from A1,
A2 and A3 respectively by joining the parallel sides
along the breadth. Then (wZbwU AvqZ‡ÿÎ A1, A2 I A3 Gi
†ÿÎdj mgvb| G‡`i ˆ`N©¨ h_vµ‡g a1, a2 I a3 †hb a1 < a2 <
a3| cÖ¯’ eivei gywo‡q A1, A2 I A3 †K c1, c2 I c3 wZbwU
wmwjÛv‡i cwiYZ Kiv n‡jv| Z‡eÑ)
a C1 will enclosed maximum volume
b C2 will enclosed maximum volume
c C2 will enclosed maximum volume
d Each of C1, C2 and C3 will enclose equal volume a

 mgvavb : A1, A2 I A3 Gi cÖ¯’ h_vµ‡g b1, b2, b3 n‡j
a1b1 = a2b2 = a3b3 = A [†ÿÎdj mgvb]
b1 =
A
a1
; b2 =
A
a2
; b3 =
A
a3
c1 Gi D”PZv = a1; cwiwa = b1
e¨vmva© r1 n‡j,
2r1 = b1
r1 =
b1
2
=
A
2a1
 AvqZb, V1 = r1
2
a1
=  
A2
42
a1
2 a1 =
A2
4a1
GKBfv‡e, c2 Gi AvqZb, V2 =
A
4a2
c3 Gi AvqZb, V3 =
A2
4a3
 c1, c2 I c3 Gi AvqZ‡bi AbycvZ =
A2
4a1
:
A2
4a2
:
A2
4a3
= V1 : V2 : V3 =
1
a1
:
1
a2
:
1
a3
= a1 < a2 < a3

1
a1
>
1
a2
>
1
a3
 V1 > V2 > V3
111. The volume of a right circular cylinder whose curved
surface area is 2640 cm2
and circumference of its base
is 66 cm, is (†Kvb wmwjÛv‡ii eµc„‡ôi †ÿÎdj 2640 cm2
I
f‚wgi cwiwa 66 cm n‡j AvqZbÑ)
a 3465 cm3
b 7720 cm3
c 13860 cm3
d 55440 cm3
c
 mgvavb : awi, f‚wgi e¨vmva© r I D”PZv h
kZ©g‡Z, 2r = 66
 r =
66
2
 r =
33  7
22
 r = 10.5 cm
Ges 2rh = 2640
 66  h = 2640
 h = 40 cm
 AvqZb = r2
h =
22
7
(10.5)2
 40 cm3
= 13860 cm3
112. A well has to be dug out that is to be 22.5 m deep and
of diameter 7m. Find the cost of plastering the inner
curved surface at 3 Tk. per sq. meter. (22.5 m Mfxi I 7
m e¨vm wewkó GKU K‚c Lbb Ki‡Z n‡e| cÖwZ eM©wgUv‡i 3 UvKv
Li‡P Gi wfZ‡ii eµ Ask cøv÷vi Ki‡Z †gvU LiP KZ?)
[www.examveda.com]
a 1465 Tk. b 1475 Tk. c 1485 Tk. d 1495 Tk. c
 mgvavb : h = 22.5 m
2r = 7 m
eµc„‡ôi †ÿÎdj = 2rh
=   2r  h
=
22
7
 7  22.5 m2
= 495 m2
 †gvU LiP = 495  3 UvKv = 1485 UvKv
113. The radius and height of a cylinder are in the ratio 5 :
7 and its volume is 4400 cm3
. Then its radius will be
(†Kvb mge„Ëf‚wgK wmwjÛv‡ii e¨vmva© I D”PZvi AbycvZ 5 : 7 Ges
AvqZb 4400 cm3
| Gi e¨vmva©Ñ)
 mgvavb : awi, e¨vmva©, r = 5x
D”PZv, h = 7x
AvqZb = r2
h
=
22
7
 (5x)2
 7x = 550x3
cÖkœg‡Z, 550 x3
= 4400
 x3
= 8
 x = 2
 e¨vmva© = 5  2 cm = 10 cm
114. The height of a right circular cylinder is 14 cm and its
curved surface is 704 sq. cm. Then its volume is : (GKwU
mge„Ëf‚wgK wmwjÛv‡ii D”PZv I eµc„‡ôi †ÿÎdj h_vµ‡g 14
cm I 704 cm2
n‡j AvqZbÑ)
a 1408 cm3
b 2816 cm3
c 5632 cm3
d 9856 cm3
b
kZ©g‡Z,
2rh = 704
 2 
22
7
 r  14 = 704
 r = 8 cm
AvqZb = r2
h
=
22
7
 (8)2
 14 cm3
= 2816 cm3
115. A closed metallic cylindrical box is 1.25 high and its
base radius is 35 cm. If the sheet metal costs 80 Tk. per
m2
, the cost of the material used in the box is (GKwU
mge„Ëf‚wgK e× avZe wmwjÛv‡ii D”PZv 1.25 m I f‚wgi e¨vmva©
35 cm| cÖwZ eM©wgUvi H avZzi g~j¨ 80 UvKv n‡j †gvU LiP KZ?)
[www.examveda.com]
a 281.60 Tk. b 290 Tk. c 340.50 Tk. d 500 Tk. a
 mgvavb : e¨vmva©, r = 35 cm = 0.35 m
mgMÖ c„‡ôi †ÿÎdj = 2r (r + h)
= 2 
22
7
 0.35  (0.35 + 1.25) m3
=
88
25
m3
 †gvU LiP =
88
25
 80 UvKv = 281.6 UvKv
116. The curved surface area of a right circular cylinder of
base radius r is obtained by multiplying its volume by
(r e¨vmva©wewkó GKwU mge„Ëf‚wgK wmwjÛv‡ii AvqZ‡bi mv‡_ KZ
¸Y Ki‡j eµc„‡ôi †ÿÎdj cvIqv hv‡e?)
a 2r b
2
r
c 2r2
d
2
r2 b
 mgvavb : r e¨vmva© wewkó wmwjÛv‡ii AvqZb = r2
h
 wb‡Y©q ¸YK =
2rh
r2
h
=
2
r
117. The ratio of total surface area to lateral surface area of
a cylinder whose radius is 20 cm and height 60 cm is
(20 cm e¨vmva© I 60 cm D”PZv wewkó GKwU wmwjÛv‡ii mgMÖc„‡ôi
†ÿÎdj I eµc„‡ôi †ÿÎd‡ji AbycvZ KZ?)
a 2 : 1 b 3 : 2 c 4 : 3 d 5 : 3 c
 mgvavb : mgMÖc„‡ôi †ÿÎdj = 2r(r + h)
 wb‡Y©q AbycvZ = 2r (r + h) : 2rh
= (r + h) : h = (20 + 60) : 60 = 80 : 60
= 8 : 6 = 4 : 3

118. Two cans have the same height equal to 21 cm. One can is
cylindrical, the diameter of whose base is 10 cm. The
other can has square base of side 10 cm. What is the
difference in their capacities? (21 cm D”PZv wewkó `yBwU
K¨v‡bi GKwU wmwjÛvivK…wZi hvi f‚wgi e¨vm 10 cm| AciwUi f‚wg
eM©vKvi hvi avi 10 cm| K¨vbØ‡qi aviY ÿgZv/AvqZ‡bi cv_©K¨
KZ?) [www.examveda.com]
a 250 cm3
b 300 cm3
c 350 cm3
d 450 cm3
d
 mgvavb : cÖ_g K¨v‡bi : D”PZv, h = 21 cm
e¨vmva©, r =
10
2
cm = 5 cm
 AvqZb = r2
h =
22
7
 52
 21 cm3
= 1650 cm3
wØZxq K¨v‡bi :
f‚wgi avi, x = 10 cm
D”PZv, h = 21 cm
AvqZb = x2
h = 102
 21 = 2100 cm3
cv_©K¨ = (2100 – 1650) cm3
= 450 cm3
119. *The diameter of the base of a cylindrical drum is 35 dm
and the height is 24 dm. It is full of kerosene. How
many tins each of size 25 cm  22 cm  35 cm can be
filled with kerosene from the drum? (†K‡ivwmb c~Y©
wmwjÛvivK…wZi GKwU Wªv‡gi f‚wgi e¨vm 35 dm Ges D”PZv 24
dm| Wªv‡g ivLv †K‡ivwmb ivL‡Z 25 cm  22 cm  35 cm gv‡ci
KZ¸‡jv wU‡bi ev· jvM‡e?) [www.examveda.com]
a 120 b 600 c 1020 d 1200 d
 mgvavb : e¨vmva©, r =
35
2
dm = 17.5 dm
= 175 cm [1 dm = 10 cm]
D”PZv, h = 24 dm = 240 cm
 Wªvg ev †K‡ivwm‡bi AvqZb = r2
h
=
22
7
 (175)2
 240 cm3
= 23100000 cm3
cÖwZwU wU‡bi AvqZb = 25 cm  22 cm  35 cm = 19250 cm3
 cÖ‡qvRbxq wU‡bi msL¨v =
23100000
19250
= 1200
120. The radius of an open cylinder is half its height and
area of the inner part is 616 sq. cms. Approximately
how many litres of milk can it contain? (gyL †Lvjv †Kvb
wmwjÛv‡ii e¨vmva© D”PZvi A‡a©K Ges wfZ‡ii mgMÖ †ÿÎdj 616
cm2
| G‡Z KZ wjUvi `ya ivLv hv‡e?)
a 1.4 b 1.53 c 1.7 d 1.9
e 2.2 b
 mgvavb : e¨vmva© =
1
2
 D”PZv
 r =
h
2
 h = 2r
GK gyL †Lvjv wmwjÛv‡ii mgMÖ †ÿÎdj = 2rh + r2
= 2r (2r) + r2
= 5 r2
cÖkœg‡Z, 5 r2
= 616
 5 
22
7
 r2
= 616  r2
= 39.2  r = 6.26 cm
 h = 2  6.26 = 12.52 cm
AvqZb = r2
h =
22
7
 6.262
 12.52 cm3
= 1541.9 cm3
= 1.5419 L [1 L = 1000 cm3
]
= 1.53 L (cÖvq)
121. The sum of the radius of the base and the height of a
solid cylinder is 37 metres. If the total surface area of
the cylinder be 1628 sq. metres, its volume is (GKwU
wb‡iU wmwjÛv‡ii f‚wgi e¨vmva© I D”PZv †hvMdj 37 m| mgMÖ
c„‡ôi †ÿÎdj 1628 m2
n‡j AvqZb KZ?)
a 3180 m3
b 4620 m3
c 5240 m3
d None of these b
 mgvavb : r + h = 37 ... ... (i)
Avevi, 2r (r + h) = 1628
 2r  37 = 1628  2 
22
7
 r  37 = 1628
 r =
7  1628
2  22  37
 r = 7
(i) bs †_‡K,
7 + h = 37
 h = 30
AvqZb = r2
h =
22
7
 72
 30 cm3
= 4620 cm3
122. *The curved surface area of a cylindrical pillar is 264 m2
and its volume is 924 m3
. Find the ratio of ts diameter
to its height. (wmwjÛvivK…wZi GKwU LyuwUi eµc„‡ôi †ÿÎdj 264
m2
I AvqZb 924 m3
| Gi f‚wgi e¨vm I D”PZvi AbycvZ wbY©q
Kiæb|) [www.examveda.com; www.examveda.com]
a 3 : 7 b 7 : 3 c 6 : 7 d 7 : 6 b
 mgvavb : e¨vm = 2r I D”PZv h
cÖkœg‡Z, 2rh = 264 ... ... (i)
r2
h = 924 ... ... (ii)
(ii)  (i) 

r2
h
2rh
=
924
264

r
2
=
7
2
 2r = 14
(i) bs †_‡K,
  14  h = 264

22
7
 14  h = 264  44h = 264  h = 6
wb‡Y©q AbycvZ = 14 : 6 = 7 : 3
123. The height of a right circular cylinder is 6 m. If three
times the sum of the areas of its two circular faces is
twice the area of the curved surface, then the radius of
its base is (GKwU mge„Ëf‚wgK wmwjÛv‡ii D”PZv 6m| Gi `yB
e„ËvKvi gy‡Li †ÿÎd‡ji mgwói 3 ¸Y I eµc„‡ôi †ÿÎd‡ji 2
¸‡Yi gvb mgvb n‡j f‚wgi e¨vmva©Ñ)
a 1 m b 2 m c 3 m d 4 m d
 mgvavb : awi, e¨vmva© I D”PZv h_vµ‡g r I h
h = 6 m|
e„ËvKvi gyL¸‡jvi †ÿÎd‡ji mgwó = 2r2
eµ c„‡ôi †ÿÎdj = 2rh
cÖkœg‡Z, 3  2r2
= 2  2rh
 6r2
= 4  r =
4
6
 h  r =
2
3
 6 m
 r = 4 m
124. The height of a closed cylinder of given volume and the
minimum surface area is (wbw`©ó AvqZ‡b ÿz`ªZg c„ôZj
wewkó GKwU wmwjÛv‡ii D”PZvÑ) [www.examveda.com]
a equal to its diameter b half of its diameter
c double of ist diameter d None of these a
 mgvavb : awi, e¨vmva© I D”PZv h_vµ‡g r I h
AvqZb, V = r2
h  h =
V
r2
c„ôZj, S = 2r2
+ 2rh
 S = 2r2
+ 2r 
V
r2  S = 2r2
+
2V
r

S †K r Gi mv‡c‡ÿ e¨eKjb K‡i cvB,
ds
dr
=
d
dr 


2r2
+
2V
r
=
d
dr
(2r2
) +
d
dr 


2V
r
= 2
d
dr
(r2
) + 2V
d
dr 


1
r
2 I 2V aªæeK

ds
dr
= 2  (2r) + 2V
d
dr
(r–1
)
= 4r + 2V (–1r–2
) = 4r –
2V
r2
S Gi b~¨bZg gv‡bi Rb¨
ds
dr
= 0
 4r –
2V
r2 = 0
 2r =
V
r2
V = 2r3
 h =
2r3
r2
 h = 2r
 D”PZv = e¨vm
125. If the radius of the base of a right circular cylinder is
halved, keeping the height same, what is the ratio of the
volume of the reduced cylinder to that of the original
one? (D”PZv mgvb †i‡L GKwU mge„Ëf‚wgK wmwjÛv‡ii e¨vmva©
A‡a©K Kiv n‡j cÖvß wmwjÛvi I gyj wmwjÛv‡ii AvqZ‡bi AbycvZ
KZ?) [www.examveda.com]
a 1 : 2 b 1 : 4 c 1 : 8 d 8 : 1 b
 mgvavb : g~j wmwjÛv‡ii AvqZb, V = r2
h
r =
r
2
I h = h n‡j,
cÖvß wmwjÛv‡ii AvqZb, V = r2
h
= 



r
2
2
 h =
1
4
r2
h
 V : V =
1
4
r2
h : r2
h = 1 : 4
126. *The radii of the bases of two cylinders are in the ratio
3 : 4 and their heights are in the ratio 4 : 3. The ratio of
their volumes is (`ywU wmwjÛv‡ii f‚wgi e¨vmv‡a©i AbycvZ 3 : 4 I
D”PZvi AbycvZ 4 : 3 n‡j AvqZ‡bi AbycvZÑ) [www.examveda.com]
a 2 : 3 b 3 : 2 c 3 : 4 d 4 : 3 c
 mgvavb : cÖ_g wmwjÛv‡ii AvqZb, V1 = r1
2
h1
wØZxq wmwjÛv‡ii AvqZb, V2 = r2
2
h2
†hLv‡b, r1 : r2 = 3 : 4
h1 : h2 = 4 : 3
V1
V2
=
r1
2
h1
r2
2
h2
=



r1
r2
2



h1
h2
=



3
4
2

4
3
=
3
4
 V1 : V2 = 3 : 4
127. If the height of a cylinder is increased by 15 percent and
the radius of its base is decreased by 10 percent then by
what percent will its curved surface area change? (hw`
GKwU wmwjÛv‡ii D”PZv 15% e„w× cvq I e¨vmva© 10% n«vm cvq, Z‡e
eµc„‡ôi †ÿÎdj kZKiv KZ cwieZ©b n‡e?) [www.examveda.com]
a 3.5 percent decrease b 3.5 percent increase
c 5 percent decrease d 5 percent increase b
 mgvavb : D”PZvi cwieZ©b, x% = 15% [e„w×]
e¨vmv‡a©i cwieZ©b, y% = – 10% [ n«vm]
eµc„‡ôi †ÿÎd‡ji kZKiv cwieZ©b = x% + y% +
xy
100
%
= 15% – 10% +
15  (–10)
100
%
= 5% – 1.5% = 3.5% [e„w×]
128. If two cylinders of equal volumes have their heights in
the ratio 2 : 3, then the ratio of their radii is (mgvb
AvqZb wewkó `ywU wmwjÛv‡ii D”PZvq AbycvZ 2 : 3 n‡j e¨vmv‡a©i
AbycvZÑ) [www.examveda.com]
a 6 : 3 b 5 : 3 c 2 : 3 d 3 : 2 d
 mgvavb : wmwjÛvi-1: e¨vmva© r1 I D”PZv h1
AvqZb = r1
2
h1
wmwjÛvi-2: e¨vmva© r2 I D”PZv h2
AvqZb = r2
2
h2
cÖkœg‡Z, r1
2
h1 = r2
2
h2




r1
r2
2
=
h2
h1

r1
r2
=
h2
h1

r1
r2
=
3
2
 r1 : r2 = 3 : 2
129. X and Y are two cylinders of the same height. The base
of X has diameter that is half the diameter of the base
of Y. If the height of X is doubled the volume of X
becomes (X I Y GKB D”PZvwewkó `ywU wmwjÛvi| X Gi f‚wgi
e¨vm Y Gi f‚wgi e¨v‡mi A‡a©K| X Gi D”PZv wØ¸Y Ki‡j, X Gi
†ÿÎdj n‡eÑ)
a equal to the volume of Y b double the volume of Y
c half the volume of Y d greater than the volume of Y c
 mgvavb : awi, Df‡qi D”PZv = h
Y Gi e¨vm = 2r
Y Gi e¨vmva© = r
X Gi e¨vm =
1
2
 2r = r
X Gi e¨vmva© =
r
2
D”PZv wØ¸Y Kivq X Gi D”PZv = 2h
 Y Gi AvqZb = r2
h
X Gi AvqZb =  



r
2
2
 2h
=
1
2
r2
h =
1
2
 Y Gi AvqZb
130. The radius of a wire is decreased to one-third and its
volume remains the same. The new length is how many
times the original length? (GKwU Zv‡ii e¨vmva© GK-Z…Zxqvsk
Kiv n‡jv wKš‘ AvqZb AcwiewZ©Z _vKj| Gi bZzb ˆ`N©¨ Avw`
ˆ`‡N©¨i KZ¸Y?)
a 1 times b 3 times c 6 times d 9 times d
 mgvavb : awi, Avw` ˆ`N©¨ = h
Avw` e¨vmva© = r
AvqZb = r2
h
ciewZ© e¨vmva© =
r
3
ciewZ© ˆ`N©¨ = h
 AvqZb = 



r
3
2
h
=
1
9
r2
h
cÖkœg‡Z,
1
9
r2
h = r2
h
 h = 9h
131. If the radius of a cylinder is decreased by 50% and the
height is increased by 50% to form a new cylinder, the
volume will be decreased by (GKwU wmwjÛv‡ii e¨vmva© 50%
n«vm I ˆ`N©¨ 50% e„w× Ki‡j Gi AvqZb n«vm cv‡eÑ)
a 0% b 25% c 62.5% d 75% c

 mgvavb : e¨vmva© cwieZ©b, x% = – 50% [n«vm]
ˆ`N©¨ cwieZ©b, y% = 50% [e„w×]
AvqZb cwieZ©b (%) = 2x% + y% +
x3
100
% +
2xy
100
% +
x2
y
100  100
%
= 2x (–50%) + 50% +
(–50)2
100
% +
2  (–50)  50
100
% +
(–50)2
 50
100  100
%
= – 62.5% [n«vm]
AvqZb cwieZ©b (%) = 2x% + y% +
x2
100
% +
2xy
100
% +
x2
y
100  100
= 2x (–50%) + 50% +



(–50)2
100
+
2  (–50)  50
100
% +
(–50)2
 50
100  100
%
= – 62.5% [n«vm]
weKí mgvavb :
awi, Avw` e¨vmva© r
Avw` D”PZv h
Avw` AvqZb r2
h
P‚ovšÍ e¨vmva© = r – r Gi 50%
= r –
50
100
r =
r
2
P‚ovšÍ D”PZv = h + h Gi 50% =
3
2
h
P‚ovšÍ AvqZb = 



r
2
2



3
2
h =
3
8
r2
h
kZKiv n«vm =
r2
h –
3
8
r2
h
r2
h
 100%
=



1 –
3
8
 100% =
5
8
 100% = 62.5%
132. Diameter of a jar cylindrical in shape is increased by
25%. By what percent must the height be decreased so
that there is no change in its volume? (wmwjÛvivK…wZi GKwU
Rv‡ii e¨vm 25% evov‡bv n‡j Gi ˆ`N©¨ KZ kZvsk Kgv‡j
AvqZ‡bi †Kvb cwieZ©b n‡e bv?) [www.examveda.com]
a 10 b 25 c 36 d 50 c
 mgvavb : e¨vm/e¨vmva© cwieZ©b, x% = 25% [e„w×]
AvqZb cwieZ©b = 0
awi, D”PZv cwieZ©b = y%
AvqZ‡bi kZKiv cwieZ©b = 2x% + y% +
x2
100
% +
2xy
100
%
+
x2
y
100  100
 2  25% + y% +
252
100
% +
2  25y
100
% +
252
y
100  100
%
 y% +
Y
2
% +
Y
16
% = – 50% –
25
4
%

25
16
y% = –
225
4
%
 y% = –1
16
25

225
4
%
y% = – 36% [n«vm]
weKí mgvavb :
awi, Avw` e¨vm 2r I D”PZv h
Avw` AvqZb = r2
h
eZ©gvb e¨vm = 2



1 +
25
100
=
5
4
 2r
eZ©gvb e¨vmva© =
5
4
r
eZ©gvb D”PZv h n‡j
AvqZb = 



5
4
r
2
h =
25
16
r2
h
cÖkœg‡Z,
25
16
r2
h = r2
h
 h =
16
25
h
 D”PZvi kZKiv n«vm =
h – h
h
 100%
=
h –
16
25
h
h
 100% =
9
25
 100% = 36%
133. A cylindrical tank of diameter 35 cm is full of water. If
11 litres of water is drawn off, the water level in the
tank will drop by (35 cm e¨vmwewkó wmwjÛvivK…wZi GKwU
U¨vsK cvwb Øviv c~Y©| 11 wjUvi cvwb †d‡j †`qv n‡j U¨vs‡K cvwbi
†j‡fj KZLvwb Kg‡e?) [www.examveda.com]
a 10
1
2
cm b 11
3
7
c 12
6
7
cm d 14 cm b
35
3
cm = 17.5 cm
11 wjUvi = 11  1000, cm3
[1 wjUvi = 1000 cm3
]
GB cwigvb cvwbi Rb¨ H cv‡Îi D”PZv h n‡j,
r2
h = 11  1000

22
7
 (17.5)2
 h = 11000
 h =
7  11000
22  (17.5)2 cm = 11
3
7
cm
134. *A well with inner diameter 8 m is dug 14 m deep. Earth
taken out of its has been evenly spread all around it to
a width of 3 m to form an embankmant. The height of
the embankment will be (8 m Af¨šÍixY e¨vm I 14 m
MfxiZv wewkó GKwU K‚c Lbb Kiv n‡jv| LbbK…Z gvwU K‚‡ci
Pvicv‡k mgvbfv‡e Qwo‡q 3 m cÖk¯Í eva ˆZwi Kiv n‡jv| euv‡ai
D”PZv KZ?) [www.examveda.com]
a 4
26
33
m b 5
26
33
m c 6
26
33
m d 7
26
33
m c
8
2
m = 4 m
MfxiZv, h = 14 m
LbbK…Z gvwUi cwigvb = r2
h =
22
7
 42
 14 m3
= 704 m3
K~‡ci f‚wgi †ÿÎdj = r2
=
22
7
 42
m2
=
352
7
m2
K‚cmn euva AÂ‡ji f‚wgi e¨vm = [8 + 2  3] m = 14 m
e¨vmva© =
14
2
m = 7 m
 †ÿÎdj =   72
m2
=
22
7
 72
m2
= 154 m2
ïaygvÎ euva AÂ‡ji †ÿÎdj =



154 –
352
7
m2
=
726
7
m2
euv‡ai D”PZv H n‡j, AvqZb =
726
7
H
cÖkœg‡Z,
726
7
H = 704
 H = 6
26
33
m

25. volume & surface areas c&add=t [www.onlinebcs.com]

25. volume & surface areas c&add=t [www.onlinebcs.com]

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