**1. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?**

A) 47500

B) 37500

C) 57500

D) 77500

**2. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is**

A) 168

B) 208

C) 306

D) 156

**3. The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is**

A) 240

B) 140

C) 40

D) 340

**4. What is the least number of squares tiles required to pave the floor of a room 15.17 m long and 9.02 m broad?**

A) 814

B) 714

C) 614

D) 713

**5. A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is?**

A) 50

B) 60

C) 70

D) 80

**6. A typist uses a sheet measuring 20cm by 30cm lengthwise.If a margin of 2 cm is left on each side and a 3 cm margin on top and bottom, then percent of the page used for typing is**

A) 64%

B) 74%

C) 84%

D) 94%

**7. A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?**

A) 56m

B) 65m

C) 34m

D) 36m

**8.A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle.**

A) 1cm

B) 2cm

C) 3cm

D) 4cm

**9. Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters.**

A) 56m

B) 16m

C) 14m

D) 76m

**10. In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements.**

A) .7%

B) 0.5%

C) 0.9%

D) 0.3%

**11. The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq. cm, find its height**

A) 6

B) 7

C) 8

D) 9

**12. Rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required?**

A) 44ft

B) 22ft

C) 54ft

D) 88ft

**13. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.**

A) B=900;H=300

B) B=300;H=900

C) B=600;H=700

D) B=500;H=900

**14. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle?**

A) 10cm

B) 20cm

C) 30cm

D) 40cm

**15. The area of a circle of radius 5 is numerically what percent its circumference?**

A) 150%

B) 250%

C) 350%

D) 450%

**Answers**

**1.(B)**

**Let length = x meters, then breadth = 0.6x**

**perimeter = 800 meters**

**=> 2[ x + 0.6x] = 800**

**=> x = 250 m**

**Length = 250m and breadth = 0.6 * 250 = 150m**

**Area = 250 * 150 = 37500**

**2.(A)**

**3.(B)**

**4.(A)**

5.(B)

6.(A)

5.(B)

6.(A)

**7.(A)**

**8. (C)**

**9. (A)**

**10. (B)**

**11. (A)**

**12. (D)**

**13. (A)**

14. (D)

15. (B)

14. (D)

15. (B)

Thank u mam….plz post trigonometry quiz post

15-B

please give with explanations madam/sir

Random answers!! 2. (C), 9. (A), 15. (B).

1-B 2-?? 3-B 4-A 5-B

Guys, Don't you think in the. Q.No 3, Option (D) should be 440 in lieu of 340 ?? Or has it been mis-typed ?

@ayushi mam plz check option of Q 2 all options r wrong

my ans is 42

Gud Evng mam/ sir….waiting for answers

Shadow!! Check the option again dear. It has 42 in (C) for Q.No 2.

Q 2

area of squre=14*14=196

area of four quater cicle=4*pie*r^2*90/360=154

required area=196-154=42 sq cm no options is in Q2

Explain 2 que plz

1-b

2-c

3-b

4-

5-b

6-a

7-

8-

9-a

10-b

11-a

12-d

13-a

14-b

15-b

scroll up i have expl

i refresh page 3 times but it still written 306(C) in Q2..

Yes ri8 bro..1st 42 option is in opinions..but after options r changed….

Refresh!

G.K. Quiz For SSC Exam

http://www.sscadda.com/2015/03/gk-quiz-for-ssc-exam.html

explain q-4 plz

plz explain q-4

length of sqr=HCF(1517 &902 )=41

then area of each tile=41*41=1681

no of tiles=1517*902/1681=814

thanks

scroll up

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = 41*41=1681

Required number of tiles = 1517*902/1681= 814

if in Q if it is asked max no of tiles then we take LCM this is main diff b/n max and min

Lcm means we get max size tile how we get max no.of tiles

no we don't get max size(max means max no of tiles not size)

for max no of tiles= LCM of nos

for min no of tiles= HCF of nos

Explain q. 13

Explain q. 13

Right option for question 3 is nt given

Right option for question 3 is nt given

please explain 5 question

6th question mae lengyth 20 lene paadegi kyaa??????

Diognal = 15*52/60=13

L+B=15*68/60=17

D^2=L^2+B^2

13^2=12^2+5^2

Area=12*5=60

thanks

take hcf

because value of hcf is smaller than Lcm

and when we will devide a value by it will provide maximum no. of tiles

Sol of 5 , area of sheet =20*30=600; area used for typing={20-4}*{30-6}=16*24=384;%=384/600*100=64%

Plz explain 9 question

speed=distance/time here distance=diagonal

v=52 m/min= 52/60 m/sec time t=15 sec

diagonal =speed*time=52/60*15=13 m

since B took the same time to cross the same field along its sides walking at the rate of 68m/min.

it means B travels along length L and breadth B

distance=L+B V=68 m/sec=68/60 m/sec t=15 sec

distance L+B=speed*time=68/60*15=17 m

Area of rectangle A= L*B

according to pythagorus therom

D^2=L^2+B^2

we can write L^2+B^2=(L+B)^2-2*L*B

=(L+B)^2-2*Area

now D^2=(L+B)^2-2*Area

13^2=17^2-2*Area

2*Area=17^2-13^2=289-169=120

Area=120/2=60 sqr m

let length of rop be r

then pie*r^2=9856

22/7*r^2=9856

r^2=9856*7/22=448*7=64*49

r= 8*7=56 m

is 2nd one ans: 42?

required area = 14*14 – (4 * 1/4 * 22/7 * 7 *7) sq cm

= 196 – 154 = 42 sq cm.

let be altitude x then base=3x

area=1/2 base*height =1/2*3x*x=(3*x^2)/2

cost=Rs. 333.18 and rate=Rs. 24.68 per hectare

Area= cost/rate=333.18/24.68=13.5 hectares

1 hectares= 10000 m^2

Area=13.5*10000=135000

(3*x^2)/2=135000

x^2=135000*2/3=90000

x=300 m

altitude x=300 m base=3x=900 m

pls explain question no. 7, 10 & 12

Explain 4,7,10 n 12 ????

Answers of the rest are as follows. 4.[A], 5.[B], 6. [A], 7. -, 8. [C], 12. [D], 13. [A], 14. [D]. I must say, after solving all but Q. No 7, what a fantastic set of Questions it was for Quantitative Aptitude !! Hats off to http://www.sscadda.com for putting quality questions before us in each section and that always. A million thanks to Ayushi Ma'am & Sandeep Sir, the persons whom I know on this importantly informative arena for the aspirants like us.

anyone plzz explain 2,4,6,7………

Q2…

area of squre=14*14=196

area of four quater(sector) cicle=4*pie*r^2*90/360=154

required area=196-154=42 sq cm no optionsis in Q2

Q4… length of sqr=HCF(1517 &902 )=41

then area of each tile=41*41=1681

no of tiles=1517*902/1681=814

Q6..

area of sheet=20*30=600 sqr cm

area used for typing=(20-4)*(30-6)=384 sqr cm

reqrd % =(384/600)*100 =64%

Q7…

length=90m breadth =50m

perimeter of rectangle=2*(90+50)=280 m

no. of pole=280/5=56

Q9…

let length of rope be r

then pie*r^2=9856

22/7*r^2=9856

r^2=9856*7/22=448*7=64*49

r= 8*7=56 m

Q10..

let length =l, breadth=b

area A=l*b

new length l'=1.05 *l, breadth b'=0.96*b

new area A'=1.05l*0.96b=1.008lb

chance in area=A'-A=1.008lb-lb

=0.008lb

error % in area=(chance in area/originalarea)*100

=(0.008lb/lb)*100= 0.8%

ans is wrong in Q 10Q12…

length l=20 ft

area=680=l*b

b=680/20= 34 ft

length of fencing=l+2b=20+2*34=20+68

=88 ft

Q2…

area of squre=14*14=196

area of four quater(sector) cicle=4*pie*r^2*90/360=154

required area=196-154=42 sq cm no optionsis in Q2

Q4… length of sqr=HCF(1517 &902 )=41

then area of each tile=41*41=1681

no of tiles=1517*902/1681=814

Q6..

area of sheet=20*30=600 sqr cm

area used for typing=(20-4)*(30-6)=384 sqr cm

reqrd % =(384/600)*100 =64%

Q7…

length=90m breadth =50m

perimeter of rectangle=2*(90+50)=280 m

no. of pole=280/5=56

Q9…

let length of rope be r

then pie*r^2=9856

22/7*r^2=9856

r^2=9856*7/22=448*7=64*49

r= 8*7=56 m

Q10..

let length =l, breadth=b

area A=l*b

new length l'=1.05 *l, breadth b'=0.96*b

new area A'=1.05l*0.96b=1.008lb

chance in area=A'-A=1.008lb-lb

=0.008lb

error % in area=(chance in area/originalarea)*100

=(0.008lb/lb)*100= 0.8%

ans is wrong in Q 10Q12…

length l=20 ft

area=680=l*b

b=680/20= 34 ft

length of fencing=l+2b=20+2*34=20+68

=88 ft.

Q2…

area of squre=14*14=196

area of four quater(sector) cicle=4*pie*r^2*90/360=154

required area=196-154=42 sq cm no optionsis in Q2

Q4… length of sqr=HCF(1517 &902 )=41

then area of each tile=41*41=1681

no of tiles=1517*902/1681=814

Q6..

area of sheet=20*30=600 sqr cm

area used for typing=(20-4)*(30-6)=384 sqr cm

reqrd % =(384/600)*100 =64%

Q7…

length=90m breadth =50m

perimeter of rectangle=2*(90+50)=280 m

no. of pole=280/5=56

Q9…

let length of rope be r

then pie*r^2=9856

22/7*r^2=9856

r^2=9856*7/22=448*7=64*49

r= 8*7=56 m

Q10..

let length =l, breadth=b

area A=l*b

new length l'=1.05 *l, breadth b'=0.96*b

new area A'=1.05l*0.96b=1.008lb

chance in area=A'-A=1.008lb-lb

=0.008lb

error % in area=(chance in area/originalarea)*100

=(0.008lb/lb)*100= 0.8%

ans is wrong in Q 10Q12…

length l=20 ft

area=680=l*b

b=680/20= 34 ft

length of fencing=l+2b=20+2*34=20+68

=88 ft..

Thank u so mch..

@!!!SHADOW!!! bro you are doin an awesome job.

hats off !!!.

@!!!SHADOW!!! bro you are doing an awesome job.

hats off.

thank u…

HI SHADOW…

I think 7th ques ans should be 52???

Thnx shadow

Thnx shadow

thanx kj

thanx :))

welcome dude.