﻿ Quantitative aptitude from R.S.Agrawal (Simple interest) - UPSC FEVER

Ans .

8500

1. Explanation :

P= Rs.6800,R= 50/3%p.a and T =9/12years=3/4 years.

S.I=(P*R*T)/100=Rs(68,000*(50/3)*(3/4)*(1/100))=Rs.8500

Ans .

37.50

1. Explanation :

2. Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years. P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a

S.I. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50.

Ans .

1625

1. Explanation :

Let sum be Rs. x then , S.I.=Rs.(x$$\frac{27}{2}$$*4*(1/100) ) = Rs.27x/50

amount = (Rs. x+$$\frac{27x}{50}$$ = $$\frac{77x}{50}$$

$$\frac{77x}{50}$$ = 2502.50

x = $$\frac{2502.50x50}{70}$$

= 1625 Hence , sum = Rs.1625.

Ans .

992

1. Explanation :

S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs.

R = ((100 x 120)/(800*3) ) % = 5%. New rate = (5 + 3)% = 8%. New S.l. = Rs. (800*8*3)/100 = Rs. 192.

New amount = Rs.(800+192) = Rs. 992.

Ans .

12,000

1. Explanation :

Let the sum borrowed be x.

Then,(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400

(3x/25 + 27x/100 + 14x / 25) = 11400

95x/100 = 11400  x = (11400*100)/95 = 12000.

Hence , sum borrowed = Rs.12,000.

Ans .

208 ,13%

1. Explanation :

S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.

S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208

Principal = Rs. (1008 - 208) = Rs. 800.

Now, P = 800, T = 2 and S.l. = 208.

Rate=(100* 208)/(800*2)% = 13

Ans .

6$$\frac{1}{4}$$% p.a

1. Explanation :

Let principal = P. Then, S.l. = P and T = 16 yrs.

Rate = (100 x P)/(P*16)% = 6$$\frac{1}{4}$$% p.a.

Ans .

6$$\frac{2}{3}$$% p.a., 6 yrs 8 months

1. Explanation :

Let sum = Rs. x. Then, S.l. = Rs. 4x/9

Let rate = R% and time = R year.

Then, (x*R*R)/100=4x/9 or $$R^2$$ =400/9 or R = 20/3 = 6 2/3.

Rate = 6$$\frac{2}{3}$$ % and Time = 6$$\frac{2}{3}$$ = 6 years 8 months.

Ans .

800

1. Explanation :

Let the sum be Rs. x Then,$$\frac{x*10*7}{100*2}$$ – $$\frac{x*12*5}{100*2}$$ = 40

$$\frac{7x}{20}$$-$$\frac{3x}{10}$$=40

x = (40 * 20) = 800.

Hence, the sum is Rs. 800.

Ans .

6000

1. Explanation :

. Let sum = P and original rate = R.

Then, [ (P*(R+2)*3)/100] – [ (P*R*3)/100] = 360.

3PR + 6P - 3PR = 36000

6P=36000

P=6000

Hence, sum = Rs. 6000

Ans .

325

1. Explanation :

.Let each Instalment be Rs. x

Then, ( x+ $$\frac{x*12*1}{100}$$ + x$$\frac{x*12*2}{100}\ ) + x = 1092 ((28x/25) + (31x/25) + x) = 1092 (28x+31x+25x)=(1092*25) x= (1092*25)/84 = Rs.325. Each instalment = Rs. 325. Ans . 650 ,900 1. Explanation : Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x). \(\frac{x*8*1}{100}$$ + $$\frac{(1550-x)*6*1}{100}$$=106

8x + 9300 –6x=10600

2x = 1300 x = 650.

Money lent at 8% = Rs. 650. Money lent at 6% = Rs. (1550 - 650) = Rs. 900.

Ans .

612

1. Explanation :

Amount = Rs [7500*$$(1+(4/100)^2$$] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112.

therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

Ans .

3109

1. Explanation :

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.

Amount = Rs'. [8000 X $$(1+(15/100))^2$$ X (1+((1/3)*15)/100)]

=Rs. [8000 * (23/20) * (23/20) * (21/20)] = Rs. 11109. .

C.I. = Rs. (11109 - 8000) = Rs. 3109.

Ans .

824.32

1. Explanation :

Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.

Amount = Rs [10000 *$$(1+(2/100))^4$$]

= Rs(10000 * (51/50) * (51/50) * (51/50) * (51/50)) = Rs. 10824.32.

C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

Ans .

2522

1. Explanation :

Principal = Rs. 16000; Time = 9 months =3 quarters;Rate = 20% per annum = 5% per quarter.

Amount = Rs. [16000 x $$(1+(5/100))^3$$] = Rs. 18522.

CJ. = Rs. (18522 - 16000) = Rs. 2522

Ans .

1261

1. Explanation :

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200

So principal=RS [100*1200]/3*5= RS 8000 Amount = Rs. 8000 x $$[1 +5/100]^3$$= Rs. 9261.

C.I. = Rs. (9261 - 8000) = Rs. 1261.

Ans .

3 yrs

1. Explanation :

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years.

Then, [ 1000 (1+ $$(10/100))^n$$ ] = 1331 or $$(11/10)^n$$ = (1331/1000) = $$(11/10)^3$$

n= 3 years.

Ans .

8% p.a.

1. Explanation :

Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.

Let the rate be R% per annum.. 'Then, [ 500 $$(1+(R/100)^2$$ ] = 583.20 or $$[1+ (R/100)]^2$$ = 5832/5000 = 11664/10000

$$[ 1+ (R/100)]^2$$ = $$(108/100)^2$$ or 1 + (R/100) = 108/100 or R = 8

So, rate = 8% p.a.

Ans .

1080

1. Explanation :

Let the sum be Rs. x.

Then, C.I. = [ x * (1 + $$( 50/(3*100))^3$$ - x ] = ((343x / 216) - x) = 127x / 216

127x /216 = 1270 or x = (1270 * 216) / 127 = 2160.

Thus, the sum is Rs. 2160 S.I. = Rs ( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080.

Ans .

Rs.63,100.

1. Explanation :

Let the sum be Rs. x. Then,

C.I. = x $$( 1 + ( 10 /100 ))^2$$- x = 21x / 100 , S.I. = (( x * 10 * 2) / 100) = x / 5

(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100

Hence, the sum is Rs.63,100.

Ans .

15%.

1. Explanation :

Let the rate be R% p.a.

then, [ 18000 ( 1 + $$( R / 100 )^2$$ ) - 18000 ] - ((18000 * R * 2) / 100 )

= 405 18000 [ ( 100 + $$(R / 100 )^2$$ / 10000) - 1 - (2R / 100 ) ]

= 405 18000[( $$(100 + R ) ^2$$ - 10000 - 200R) / 10000 ]

= 405 9$$R^2$$ / 5 = 405 $$R^2$$ =((405 * 5 ) / 9) = 225

R = 15. Rate = 15%.

Ans .

Rs.676 and Rs.625.

1. Explanation :

Let the two parts be Rs. x and Rs. (1301 - x).

x$$(1+4/100)^7$$ =$$(1301-x)(1+4/100)^9$$ x/(1301-x)=$$(1+4/100)^2$$=(26/25*26/25) 625x=676(1301-x) 1301x

=676*1301 x=676. So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625.

Ans .

Rs.5400

1. Explanation :

S.I on rs.7350 for 1 year=rs.(8575-7350)=rs.1225.

Rate=(100*1225/7350*1)%=16 2/3%

Let the sum be rs.x.then, X$$(1+50/3*100)^2$$=7350 X*7/6*7/6=7350

X=(7350*36/49)=5400. Sum=rs.5400

Ans .

Rs.4460.

1. Explanation :

. Let the sum be rs.P.then P$$(1+R/100)^3$$=6690…(i) and P$$(1+R/100)^6$$=10035…(ii)

On dividing,we get $$(1+R/100)^3$$=10025/6690=3/2.

Substituting this value in (i),we get: P*3/2=6690 or P=(6690*2/3)=4460

Hence,the sum is rs.4460.

Ans .

45 years

1. Explanation :

$$P(1+R/100)^15$$=2P $$(1+R/100)^15$$=2P/P=2

LET $$P(1+R/100)^n$$=8P

$$(1+R/100)^n$$=8=23=$$((1+R/100)^15))^3$$

[USING (I)]

$$(1+R/100)^N$$=$$(1+R/100)^45$$ n=45.

Thus,the required time=45 years

Ans .

Rs.3430

1. Explanation :

Let each installment beRs.x.

Then,(P.W. of Rs.x due 1 year hence)+(P>W of Rs.x due 2 years hence)+(P.W of Rs. X due 3years hence)=7620.

x/(1+(50/3*100))+ x/$$(1+(50/3*100))^2$$ + x/$$(1+(50/3*100))^3$$=7620

(6x/7)+(936x/49)+(216x/343)=7620.

294x+252x+216x=7620*343.

x=(7620*343/762)=3430.

Amount of each installment=Rs.3430.

Ans .

3,-3,-1

1. Explanation :

(1) let LOG(3, 27)=$$3^3$$ or n=3.

ie, $$log_3$$(27)= 3.

(2) Let $$log_7$$(1\343) = n.

Then ,7n =1/343=1/73

n = -3.ie,log7(1\343)= -3.

(3) let $$log_100$$(0.01) = n.

Then,. (100) = 0.01 = 1 /100=100 -1 0r n=-1

Ans .

0,0,16

1. Explanation :

solution: i) we know that $$log_a$$ 1=0 ,so $$log_7$$ 1=0 . ii) we know that $$log_a$$a=1,so $$\log_{34} 34$$ =0. iii) We know that $$a^{\log_{6} x}$$ =x.

now $$36^{\log_{6} 4}$$=$$6^{2^{\log_{6} 4}}$$ =$$6^{\log_{6} 16}$$=16.

Ans .

32

1. Explanation :

$$\log_{\sqrt{8}} x$$=10/3

x=$$\sqrt{8}^{10/3}$$ =$$2^{\frac{2}{3} ^{\frac{10}{3}}}$$=$$2^{\frac{2}{3} * \frac{10}{3}}$$=25=32.

Ans .

2/3,5/6

1. Explanation :

$$\log_{5} 3$$* $$\log_{27} 25$$=log 3/log 5*log 25/log 27 =(log 3 /log 5) * log$$5^2$$ *log$$3^3$$

=(log 3/log 5)*(2log 5 / 3(log 3)

=2/3

(ii)Let $$\log_{9} 27$$=n

Then,$$9 ^n$$ =27 $$3^{2n}$$ =$$3^3$$ 2n=3 n=3/2

Again, let $$\log_{27} 9$$=m

Then,$$27^m$$ =9 $$3^{3m}$$ =$$3 ^2$$ 3m=2 m=2/3

$$\log_{9} 27$$- $$\log_{27} 9$$=(n-m)=(3/2-2/3)=5/6

Ans .

log2

1. Explanation :

log 75/16-2 log 5/9+log 32/243

= log 75/16-log(5/9)2+log32/243

= log 75/16-log25/81+log 32/243

= log(75/16*32/243*81/25)=log 2

Ans .

7/2

1. Explanation :

$$\log_{10} 3$$+$$\log_{10} (4x+1)$$=$$\log_{10} (x+1)$$+1

$$\log_{10} 3$$+$$\log_{10} (4x+1)$$=$$\log_{10} (x+1)$$+$$\log_{10} (x+1)$$+$$\log_{10} 10$$

$$\log_{10} (3(4x+1))$$=$$\log_{10} (10(x+1))$$ =3(4x+1)=10(x+1)=12x+3 =10x+10 =2x=7=x=7/2

Ans .

2

1. Explanation :

$$\log_{xyz} (xy)$$ + $$\log_{xyz} (yz)$$ + $$\log_{xyz} (zx)$$

=$$\log_{xyz} (xy*yz*zx)$$=$$\log_{xyz} {(xyz)^2}$$ 2$$\log_{xyz} (xyz)$$ =2*1=2

Ans .

1.69897

1. Explanation :

log 50=log (100/2)=log 100-log 2=2-0.30103=1.69897.

Ans .

1.39,0.65

1. Explanation :

i) log 25=log(100/4)=log 100-log 4=2-2log 2=(2-2*.3010)=1.398.

ii) log 4.5=log(9/2)=log 9-log 2=2log 3-log 2=(2*0.4771-.3010)=0.6532

Ans .

17

1. Explanation :

log $$2^{56}$$ =56 log2=(56*0.30103)=16.85768. Its characteristics is 16. Hence,the number of digits in $$2^{56}$$ is 17

Ans .

120 $$m^2$$

1. Explanation :

Other side = $${17}^ 2$$- $$15^2{^{(1/2)}}$$ = $$(289- 225)^{1/2}$$ = $${64}^{1/2}$$ = 8 m.

Area = (15 x 8) $$m^2$$ = 120 $$m^2$$

Ans .

33/3,50

So, 2x * 3x = 5000/3 <=> $$x^2$$ = 2500/9 <=> x = 50/3

therefore Length = 2x = (100/3) m = 33(1/3) m and Breadth = 3x = 3(50/3) m = 50m.

Ans .

Rs. 1934.40.

1. Explanation :

Area of the carpet = Area of the room = (13 * 9)$$m^2$$ = 117 $$m^2$$. Length of the carpet = (area/width) = 117 *(4/3) m = 156 m. Therefore Cost of carpeting = Rs. (156 * 12.40) = Rs. 1934.40.

Ans .

120 $$cm^2$$

1. Explanation :

Let length = x and breadth = y. Then,

2 (x + y) = 46 or x + y = 23 and $$x^2$$ + $$y^2$$ = $$17^2$$ = 289.

Now, $$(x+y)^2$$ = $$23^2$$ <=> ( $$x^2$$ + $$y^2$$ ) + 2xy = 529 <=> 289 + 2xy = 529 xy=120

Area = xy = 120 $$cm^2$$ .

Ans .

20 cm

1. Explanation :

Let breadth = x. Then, length = 2x. Then,

(2x - 5) (x + 5) - 2x * x = 75 <=> 5x - 25 = 75 <=> x = 20.

Length of the rectangle = 20 cm.

Ans .

0.8%.

1. Explanation :

Let x and y be the sides of the rectangle. Then, Correct area = xy.

Calculated area = (105/100)*x * (96/100)*y = (504/500 )(xy)

Error In measurement = (504/500)xy- xy = (4/500)xy

Error % = [(4/500)xy *(1/xy) *100] % = (4/5) % = 0.8%.

Ans .

Rs. 680

1. Explanation :

Area of the plot = (110 x 65) $$m^2$$ = 7150 $$m^2$$

Area of the plot excluding the path = [(110 - 5) * (65 - 5)] $$m^2$$ = 6300 $$m^2$$ .

Area of the path = (7150 - 6300) $$m^2$$ = 850 v.

Cost of gravelling the path = Rs.850 * (80/100)= Rs. 680

Ans .

24 cm

1. Explanation :

Side of first square = (40/4) = 10 cm;

Side of second square = (32/4)cm = 8 cm.

Area of third square = [(10) 2 - (8) 2] $$cm^2$$ = (100 - 64) $$cm^2$$ = 36 $$cm^2$$

Side of third square = (36)(1/2) cm = 6 cm.

Required perimeter = (6 x 4) cm = 24 cm.

Ans .

176

1. Explanation :

Area of the room = (544 x 374) $$cm^2$$.

Size of largest square tile = H.C.F. of 544 cm and 374 cm = 34 cm.

Area of 1 tile = (34 x 34) $$cm^2$$.

Number of tiles required =(544*374)/(34*34)=176

Ans .

7.22 $$m^2$$

1. Explanation :

Area of the square = (1/2)* $$(diagonal)^2$$ = [(1/2)*3.8*3.8 ]$$m^2$$ = 7.22 $$m^2$$ .

Ans .

4 : 25.

1. Explanation :

Let the diagonals of the squares be 2x and 5x respectively.

Ratio of their areas = (1/2)*$$(2x) ^2$$ :(1/2)*$$(5x)^ 2$$ = 4$$x^2$$ : 25$$x^2$$ = 4 : 25.

Ans .

56.25%

1. Explanation :

Let each side of the square be a. Then, area = $$a^2$$.

New side =(125a/100) =(5a/4). New area = $$(5a/4) ^2$$ =(25$$a^2$$)/16.

Increase in area = ((25 $$a^2$$)/16)-$$a^2$$ =(9$$a^2$$)/16.

Increase% = [((9$$a^2$$)/16)*(1/$$a^2$$)*100] % = 56.25%.

Ans .

50 cm

1. Explanation :

Let x and y be the length and breadth of the rectangle respectively.

Then, x - 4 = y + 3 or x - y = 7 ----(i)

Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)

(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)

Solving (i) and (ii), we get x = 16 and y = 9.

Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Ans .

l=9,b=6,h=6

1. Explanation :

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate/m2)=(270/5)m2=54m2.

x* (3x/2) = 54 <=> $$x^2$$= (54*2/3) = 36 <=> x = 6.

So, breadth = 6 m and length =(3/2)*6 = 9 m.

Now, papered area = (1720/10)$$m^2$$ = 172 $$m^2$$.

Area of 1 door and 2 windows = 8 $$m^2$$.

Total area of 4 walls = (172 + 8)$$m^2$$ = 180 $$m^2$$

2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.

Ans .

84 $$cm^2$$..

1. Explanation :

Let a = 13, b = 14 and c = 15. Then, S = (1/2)(a + b + c) = 21.

(s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = $$(s(s- a) (s - b)(s - c))^{1/2}$$ = $$(21 *8 * 7*6)^{1/2}$$ = 84 $$cm^2$$.

Ans .

30$$cm^2$$

1. Explanation :

Height of the triangle = [$$13^2$$ - $$12^{2 ^{1/2}}$$] cm = $$25^{1/2}$$ cm = 5 cm.

Its area = (1/2)* Base * Height = ((1/2)*12 * 5) $$cm^2$$ = 30$$cm^2$$ .

Ans .

Base = 900 m and Altitude = 300 m

1. Explanation :

Area of the field = Total cost/rate = (333.18/25.6)hectares = 13.5 hectares

(13.5 x 10000) $$m^2$$ = 135000 $$m^2$$

Let altitude = x metres and base = 3x metres.

Then, (1/2)* 3x * x = 135000 <=>$$x^2$$ = 90000 <=>x = 300.

Base = 900 m and Altitude = 300 m.

Ans .

60 $$cm^2$$

1. Explanation :

Let ABC be the isosceles triangle and AD be the altitude.

Let AB = AC = x. Then, BC = (32 - 2x).

Since, in an isosceles triangle, the altitude bisects the base,

so BD = DC = (16 - x).

In triangle ADC, $$AC^2$$= AD +$$DC^2$$ =>$$x^2$$ =(82)+$$(16-x) ^2$$

=>32x = 320 =>x= 10.

BC = (32- 2x) = (32 - 20) cm = 12 cm.

Hence, required area = ((1/2)x*BC * AD) = ((1/2)*12 *10)$$cm^2$$ = 60 $$cm^2$$

Ans .

4.5 cm.

1. Explanation :

Area of the triangle = ($$\sqrt{3}$$/4) x (3$$\sqrt{3}$$)2 = 27$$\sqrt{3}$$. Let the height be h.

Then, (1/2) x 3$$\sqrt{3}$$ x h = (27$$\sqrt{3}$$/4) X(2/$$\sqrt{3}$$) = 4.5 cm.

Ans .

16 : 9.

1. Explanation :

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.

Then,

((1/2) X x X 3h)/(1/2) X y X 4h) =4/3 => x/y =(4/3 X 4/3)=16/9

Required ratio = 16 : 9.

Ans .

6 cm

1. Explanation :

Let the height of the parallelogram be x. cm. Then, base = (2x) cm.

2x X x =72 2$$x^2$$ = 72 $$x^2$$ =36 x=6

Hence, height of the parallelogram = 6 cm.

Ans .

384 $$cm^2$$

1. Explanation :

Let other diagonal = 2x cm.

Since diagonals of a rhombus bisect each other at right angles, we have:

$$20^2$$ = $$12^2$$ + $$x^2$$ x= $$\sqrt{20^2-{12^2}}$$= 256= 16 cm.

So, other diagonal = 32 cm.

Area of rhombus = (1/2) x (Product of diagonals) = ((1/2)x 24 x 32) $$cm^2$$ = 384 $$cm^2$$

Ans .

27 cm and 23 cm

1. Explanation :

Let the two parallel sides of the trapezium be a em and b em.

Then, a - b = 4

And, (1/2) x (a + b) x 19 = 475 --> (a + b) =((475 x 2)/19) --> a + b = 50

Solving (i) and (ii), we get: a = 27, b = 23.

So, the two parallel sides are 27 cm and 23 cm

Ans .

paste right option

1. Explanation :

Clearly, the cow will graze a circular field of area 9856 sq. metres and radius equal to the length of the rope. Let the length of the rope be R metres. Then, $$\prod R^2$$= (9856 X (7/22)) = 3136 --> R = 56. Length of the rope = 56 m.

Ans .

Rs. 5808

1. Explanation :

Area = (13.86 x 10000) $$m^2$$ = 138600 $$m^2$$ .

($$R^2$$ = 138600 ($$R^2$$ = (138600 x (7/22)) R = 210 m.

Circumference = 2$$\prod$$R = (2 x (22/7) x 210) m = 1320 m.

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

Ans .

250.

1. Explanation :

Distance to be covered in 1 min. = $$\frac{66 * 1000}{60}$$ m = 1100 m.

Circumference of the wheel = (2 x (22/7) x 0.70) m = 4.4 m.

Number of revolutions per min. =(1100/4.4) = 250.

Ans .

14 m.

1. Explanation :

Distance covered in one revolution =$$\frac{88 X 1000}{1000}$$= 88m.

2$$\prod$$R = 88 --> 2 x (22/7) x R = 88 --> R = 88 x (7/44) = 14 m.

Ans .

84 m

1. Explanation :

Let inner radius be r metres. Then, 2$$\prod$$r = 440 --> r = (440 x (7/44))= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

Ans .

4 m

1. Explanation :

Let the inner and outer radii be r and R metres.

Then 2$$\prod$$r = (352/7) -->r =((352/7) X (7/22) X (1/2))=8m.

2$$\prod$$R=(528/7) --> R=((528/7) X (7/22) X (1/2))= 12m.

Width of the ring = (R - r) = (12 - 8) m = 4 m.

Ans .

3 cm.

1. Explanation :

Let the radius of the circle be r cm. Then,

$$\frac{\prod r^{2}\theta }{360}$$=(66/7) --> (22/7) X $$r^2$$ X(120/360)= (66/7)

$$r^2$$=((66/7) X (7/22) X 3) --> r=3. Hence, radius = 3 cm.

Ans .

1 : 2.

1. Explanation :

Radius of circum circle= ($$\sqrt{2}$$x/2) =(x/$$\sqrt{2}$$)

Required ratio = (($$\frac{\prod r^2}{4}$$ ) : (($$\frac{\prod r^2}{2}$$ ) = (1/4) : (1/2) = 1 : 2.

Ans .

75%

1. Explanation :

Original area=$$\prod {r/2}^2$$= and new area= $$\prod {r/2}^2$$= (($$\frac{\prod r^2}{4}$$ )

Decrease in area =((3$$\frac{\prod r^2}{4}$$ ) X (1/$$\prod {r/2}^2$$ X 100) % = 75%

Ans .

868 $$cm^2$$

1. Explanation :

Volume = (16 x 14 x 7) $$m^3$$ = 1568 $$m^3$$ Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] $$cm^2$$ = (2 x 434) $$cm^2$$= 868 $$cm^2$$.

Ans .

17 m.

1. Explanation :

Length of longest pole = Length of the diagonal of the room

= $$\sqrt{12^{2}+8^{2}+9^{2}}$$= .$$\sqrt{289}$$= 17 m.

Ans .

40 cm

1. Explanation :

Let the breadth of the wall be x metres.

Then, Height = 5x metres and Length = 40x metres.

x * 5x * 40x = 12.8 --> $$x^3$$=12.8/200 = 128/2000 = 64/1000

So, x = (4/10) m =((4/10)*100)cm = 40 cm

Ans .

45000.

1. Explanation :

Volume of the wall = (2400 x 800 x 60) cu. cm.

Volume of bricks = 90% of the volume of the wall

=((90/100)*2400 *800 * 60)cu.cm.

Volume of 1 brick = (24 x 12 x 8) cu. cm.

Number of bricks=(90/100)*(2400*800*60)/(24*12*8)=45000.

Ans .

96min

1. Explanation :

Volume required in the tank = (200 x 150 x 2) $$x^3$$ = 60000 $$x^3$$

Length of water column flown in1 min =(20*1000)/60 m =1000/3 m

Volume flown per minute = 1.5 * 1.25 * (1000/3) $$x^3$$ = 625 $$x^3$$

Required time = (60000/625)min = 96min

Ans .

8.04 kg.

1. Explanation :

Volume of the metal used in the box = External Volume - Internal Volume

= [(50 * 40 * 23) - (44 * 34 * 20)]$$cm^3$$

= 16080 $$cm^3$$

Weight of the metal =((16080*0.5)/1000) kg = 8.04 kg.

Ans .

216 $$cm^2$$

1. Explanation :

Let the edge of the cube be a.

$$\sqrt{3}$$a = 6../3 _ a = 6.

So,Volume = $$a^3$$ = (6 x 6 x 6) cm3 = 216 $$cm^3$$

Surface area = 6$$a^2$$ = (6 x 6 x 6) $$cm^2$$== 216 $$cm^2$$

Ans .

4913 $$cm^3$$.

1. Explanation :

Let the edge of the cube bea. Then,

6$$a^2$$ = 1734 $$a^2$$ = 289 => a = 17 cm.

Volume = $$a^3$$= (17)3 $$cm^3$$= 4913 $$cm^3$$.

Ans .

40.

1. Explanation :

Volume of the block = (6 x 12 x 15)$$cm^3$$. = 1080$$cm^3$$.

Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm.

Volume of this cube = (3 x 3 x 3) $$cm^3$$. = 27 $$cm^3$$..

Number of cubes = 1080/27 = 40.

Ans .

11.25 cm

1. Explanation :

Increase in volume = Volume of the cube = (15 x 15 x 15)$$cm^3$$.

Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.

Ans .

486 $$cm^2$$

1. Explanation :

Volume of new cube = ($$1^3$$ +$$6^3$$ + $$8^3$$) cm = 729 $$cm^3$$

Edge of new cube = $$\sqrt[3]{729}$$cm = 9 cm.

Surface area of the new cube = (6 x 9 x 9) $$cm^2$$ = 486 $$cm^2$$.

Ans .

125%

1. Explanation :

Let original length of each edge = a.

Then, original surface area = 6$$a$$ .

New edge = (150% of a) = (150a/100) = 3a/2

New surface area = 6x $${(3a/2)}^2$$ = 27 $$a^2$$/2

Increase percent in surface area =( $$\frac{15a^2}{2}$$ x $$\frac{1}{6a^2}$$ x 100)% = 125%

Ans .

1:9

1. Explanation :

Let their edges be a and b. Then,

$$a^3$$./$$b^3$$= 1/27 (or) $${(a/b)}^3$$ = $${(1/3)}^3$$ (or) (a/b) = (1/3).

Ratio of their surface area = 6$$a^2$$/6$$b^2$$ = $$a^2$$/$$b^2$$ = $${(a/b)}^2$$ = 1/9, i.e. 1:9.

Ans .

957 $$cm^2$$

1. Explanation :

Volume = $$\prod r^{2}$$h = ((22/7)x(7/2)x(7/2)x40) = 1540 $$cm^3$$. .

Curved surface area = 2$$\prod$$rh = (2x(22/7)x(7/2)x40)= 880 $$cm^2$$.

Total surface area = 2$$\prod$$rh + 2$$\prod r^{2}$$ = 2$$\prod$$r (h + r)

= (2 x (22/7) x (7/2) x (40+3.5)) $$cm^2$$= 957 $$cm^2$$.

Ans .

12 m

1. Explanation :

Let the depth of the tank be h metres. Then,

$$\prod$$ x $$7^2$$ x h = 1848 --> h = (1848 x (7/22) x (1/49) = 12 m

Ans .

112 m.

1. Explanation :

Let the length of the wire be h metres. Then,

$$\prod {(0.50/(2 x 100))}^2$$ x h = 2.2/1000

h = ( (2.2/1000) x (100 x 100)/(0.25 x 0.25) x (7/22) ) = 112 m.

Ans .

400

1. Explanation :

Volume of 1 rod = (( 22/7) x (1/100) x (1/100) x 7 ) cu.m = 11/5000 cu.m

Volume of iron = 0.88 cu. m.

Number of rods = (0.88 x 5000/11) = 400.

Ans .

2.5

1. Explanation :

Let the radii of the cylinders be 3x, 5x and their heights be 2y, 3y respectively. Then

Ratio of their curved surface area = $$\frac{2\prod X 3x X 2y}{2\prod X 5x X 3y}$$ = 2/5 = 2.5

Ans .

26.4 kg.

1. Explanation :

Inner radius = (3/2) cm = 1.5 cm, Outer radius = (1.5 + 1) = 2.5 cm.

Volume of iron = [$$\prod$$x (2.5)2 x 100 -$$\prod$$ x (1.5)2 x 100] $$cm^{3}$$

= (22/7) x 100 x [(2.5)2 - (1.5)2] $$cm^{3}$$

= (8800/7) $$cm^{3}$$

Weight of the pipe = ((8800/7) x (21/1000))kg = 26.4 kg.

Ans .

35cm

1. Explanation :

Here, r = 21 cm and h = 28 cm.

Slant height, l = $$\sqrt{r^{2}+h^{2}}$$ = $$\sqrt{21^{2}+28^{2}}$$ =$$\sqrt{1225}$$ = 35cm

Ans .

440 m.

1. Explanation :

Here, r = 7m and h = 24 m.

So,l = $$\sqrt{r^{2}+h^{2}}$$ =$$\sqrt{7^{2}+24^{2}}$$ =$$\sqrt{625}$$ = 25 m.

Area of canvas =$$\prod$$rl=((22/7)*7*25) $$m^{2}$$= 550 $$m^{2}$$.

Length of canvas = (Area/Width) = (550/1.25) m = 440 m.

Ans .

9 : 32

1. Explanation :

Let the radii of their bases be r and R and their heights be h and 2h respectively.

Then,(2$$\prod$$r/2$$\prod$$R)=(3/4)--> R=(4/3)r.

Ratio of volumes = (((1/3)$$\prod$$ $$r^{2}$$h)/((1/3)$$\prod$$$${4/3r}^2$$(2h)))=9 : 32.

Ans .

9 : 8.

1. Explanation :

Let the radii of the cylinder and the cone be 3r and 4r and their heights be 2h and

3h respectively.

$$\frac{Volume of cylinder}{Volume of cone}$$ =$$\frac{\prod3 r^{2}* 2h}{(1/3)\prod r^{2}* 3h}$$= 9/8 = 9 : 8.

Ans .

24 cm

1. Explanation :

Volume of the liquid in the cylindrical vessel

= Volume of the conical vessel

= ((1/3)* (22/7)* 12 * 12 * 50) )$$cm^{3}$$ = (22 *4 *12 * 50)/7 $$cm^{3}$$

Let the height of the liquid in the vessel be h.

Then (22/7)*10*10*h =(22*4*12*50)/7 or h = (4*12*50)/100 = 24 cm

Ans .

4851 $$cm^{3}$$,1386 $$cm^{2}$$

1. Explanation :

Volume = (4/3)$$r^{3}$$ =(4/3)*(22/7)*(21/2)*(21/2)*(21/2) $$cm^{3}$$ = 4851 $$cm^{3}$$.

Surface area = 4$$\prod r^{2}$$ =(4*(22/7)*(21/2)*(21/2)) $$cm^{2}$$ = 1386 $$cm^{2}$$

Ans .

237.5%, 125%.

1. Explanation :

Original volume = (4/3)$$\prod R^{3}$$, New volume = (4/3)$$\prod {3R/2}^{3}$$ =($$\prod R^{3}$$/2)

Increase % in volume=((19/6)$$\prod R^{3}$$)*(3/4$$\prod R^{3}$$)*100))% = 237.5%

Original surface area =4$$\prod R^{2}$$. New surface area = 4$$\prod {3R/2}^{2}$$=9($$\prod R^{2}$$)

Increase % in surface area =(5$$\prod R^{2}$$/4$$\prod R^{2}$$) * 100) % = 125%.

Ans .

1728.7

1. Explanation :

Volume of larger sphere = (4/3)$$\prod$$*6*6*6) $$cm^{3}$$ = 288$$\prod$$ $$cm^{3}$$

Volume of 1 small lead ball = ((4/3)$$\prod$$*(1/2)*(1/2)*(1/2)) $$cm^{3}$$ = $$\prod$$/6 $$cm^{3}$$

Number of lead balls = (288$$\prod$$*(6/$$\prod$$)) = 1728.7

Ans .

1792

1. Explanation :

Volume of cylinder = ($$\prod$$ x 6 x 6 x 28 ) $$cm^{3}$$= ( 9$$\prod$$/16) $$cm^{3}$$.

Number of bullet =$$\frac{Volume of cylinder}{Volume of each bullet}$$ = [(36 x 28)$$\prod$$ x 16] /9$$\prod$$ = 1792.

Ans .

243m

1. Explanation :

Volume of sphere = ((4$$\prod$$/3) x 9 x 9 x 9 ) $$cm^{3}$$ = 972$$\prod$$$$cm^{3}$$

Volume of sphere = ($$\prod$$ x 0.2 x 0.2 x h ) $$cm^{3}$$

972$$\prod$$= $$\prod$$ x (2/10) x (2/10) x h --> h = (972 x 5 x 5 )cm = [( 972 x 5 x5 )/100 ] m

= 243m

Ans .

4.2.cm

1. Explanation :

Volume of sphere = Volume of 2 cones

= ($$\frac{1}{3}\prod$$x (2.102) x 4.1 + $$\frac{1}{3} \prod x {2.1}^2$$x 4.3)

Let the radius of sphere be R

(4/3)$$\prod R^{3}$$ = (1/3)$$\prod {2.1}^{3}$$ or R = 2.1cm

Hence , diameter of the sphere = 4.2.cm

Ans .

1:2

1. Explanation :

Let radius of each be R and height of the cone be H.

Then, (4/3) $$\prod R^{3}$$ = (1/3) $$\prod R^{2}$$H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2

Required ratio = 1:2.

Ans .

1039.5 $$cm^{2}$$

1. Explanation :

Volume = (2 $$\prod r^{3}$$/3) = ((2/3) x (22/7) x (21/2) x (21/2) x (21/2))$$cm^{3}$$

= 2425.5 $$cm^{3}$$

Curved surface area = 2$$\prod r^{3}$$ = (2 x (22/7) x (21/2) x (21/2))$$cm^{2}$$

=693 $$cm^{2}$$

Total surface area = 3$$\prod r^{3}$$ = (3 x (22/7) x (21/2) x (21/2))$$cm^{2}$$

= 1039.5 $$cm^{2}$$.

Ans .

54

1. Explanation :

Volume of bowl = ((2($$\prod$$/3) x 9 x 9 x 9 ) $$cm^{3}$$= 486($$\prod$$$$cm^{3}$$.

Volume of 1 bottle = ($$\prod$$ x (3/2) x (3/2) x 4 ) $$cm^{3}$$ = 9($$\prod$$$$cm^{3}$$

Number of bottles = (486($$\prod$$/9($$\prod$$) = 54.

Ans .

1:2:3

1. Explanation :

Let R be the radius of each

Height of the hemisphere = Its radius = R.

Height of each = R.

Ratio of volumes = (1/3)$$\prod R^{2}$$ x R : (2/3)$$\prod R^{3}$$ : $$\prod R^{2}$$ x R = 1:2:3

Ans .

4 min. 3 s

1. Explanation :

Clearly, B covers 28 m in 7 seconds.

B's time over the course = (278 x 1000) sec = 250 seconds.

A's time over the course = (250 - 7-) sec = 243 sec = 4 min. 3 s

Ans .

196 m

1. Explanation :

Ratio of the rates of A and B = 7/4 : 1 = 7 : 4.

So, in a race of 7 m, A gains 3m over B.

3 m are gained by A in a race of 7 m.

84 m are gained by A in a race of (7/3 x 84) m = 196 m.

Winning post must be 196 m away from the starting point

Ans .

50 m

1. Explanation :

Clearly, A beats B by 10 sec.

Distance covered by B in 10 sec. = ($$\frac{1000 }{200}$$x 10 )m = 50 m.

Therefore A beats B by 50 metres

Ans .

5.76 km/hr

1. Explanation :

Time taken by A to cover 100 m =(60 X 60 / 8000) x 100 sec = 45 sec.

B covers (100 - 4) m = 96 m in (45 + 15) sec = 60 sec.

B's speed =$$\frac{96 *60 * 60}{60 *1000}$$ km/hr = 5.76 km/hr.

Ans .

936 m

1. Explanation :

While A covers 1000 m, B covers (1000 - 40) m = 960 m and

C covers (1000 - 64) m or 936 m.

When B covers 960 m, C covers 936 m.

Ans .

8 points

1. Explanation :

A: B = 80 : 75, A : C = 80 : 65.

B/C = ( B/ A * A/C) = (75 / 80 * 80 / 65) = 15/13 = 60 /52 = 60: 5

Therfore ,In a game of 60, B can give C 8 points

Ans .

Tuesday

1. Explanation :

16th July, 1776 = (1775 years + Period from 1st Jan., 1776 to 16th July,1776)

Counting of odd days :

1600 years have 0 odd day. 100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years)

= [(18 x 2) + (57 x 1)] odd days = 93 odd days

= (13 weeks + 2 days) = 2 odd days.

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 +16 = 198days

= (28 weeks + 2 days) =2days

Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.

Ans .

Saturday

1. Explanation :

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th

Counting of odd days:

1600 years have 0 odd day. 300 years have 1 odd day.

47 years = (11 leap years + 36 ordinary years)

= [(11 x 2) + (36 x 1») odd days = 58 odd days = 2 odd days.

Jan. Feb. March April May June July Aug.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15

= 227 days = (32 weeks + 3 days) = 3,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was 'Saturday'.

Ans .

Sunday

1. Explanation :

16th April, 2000 = (1999 years + Period from 1st Jan., 2000 to 16thA'

Counting of odd days:

1600 years have 0 odd day. 300 years have 1 odd day.

99 years = (24 leap years + 75 ordinary years)

= [(24 x 2) + (75 x 1)] odd days = 123 odd days

= (17 weeks + 4 days) = 4 odd days.

Jan. Feb. March April

31 + 29 + 31 + 16 = 107 days = (15 weeks + 2 days) = 2 odd,

Total number of odd days = (0 + 1 + 4 + 2) odd days = 7 odd days = 0 oddday.

Hence, the required day was 'Sunday'.

Ans .

5th, 12th, 19th and 26th

1. Explanation :

Let us find the day on 1st July, 2004.

2000 years have 0 odd day. 3 ordinary years have 3 odd days.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 1

= 183 days = (26 weeks + 1 day) = 1 t .

Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. '

1st July 2004 was 'Thursday',-,-

Thus, 1st Monday in July 2004 _as on 5th July.

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.

Ans .

47*1/2°

1. Explanation :

angle traced by the hour hand in 12 hours = 360°

Angle traced by it in three hours 25 min (ie) 41/12 hrs=(360*41/12*12)°

=102*1/2°

angle traced by minute hand in 60 min. = 360°.

Angle traced by it in 25 min. = (360 X 25 )/60= 150°

Required angle = 1500 – 102*1/2°= 47*1/2°

Ans .

120/11 min. past 2

1. Explanation :

At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they

are 10 min spaces apart.

To be together, the minute hand must gain 10 minutes over the hour hand.

Now, 55 minutes are gained by it in 60 min.

10 minutes will be gained in (60 x 10)/55 min. = 120/11 min.

The hands will coincide at 120/11 min. past 2.

Ans .

paste right option

1. Explanation :

At 4 o'clock, the minute hand will be 20 min. spaces behind the hour hand,

Now, when the two hands are at right angles, they are 15 min. spaces apart. So,

they are at right angles in following two cases.

Case I. When minute hand is 15 min. spaces behind the hour hand:

In this case min. hand will have to gain (20 - 15) = 5 minute spaces. 55 min. spaces are gained by it in 60 min.

5 min spaces will be gained by it in 60*5/55 min=60/11min.

They are at right angles at 60/11min. past 4.

Case II. When the minute hand is 15 min. spaces ahead of the hour hand:

To be in this position, the minute hand will have to gain (20 + 15) = 35 minute spa' 55 min. spaces are gained in 60 min.

35 min spaces are gained in (60 x 35)/55 min =40/11

They are at right angles at 40/11 min. past 4

Ans .

120/11 min

1. Explanation :

At 8 o'clock, the hour hand is at 8 and the minute hand is at 12, i.e. the two

hands_ are 20 min. spaces apart.

To be in the same straight line but not together they will be 30 minute spaces apart.

So, the minute hand will have to gain (30 - 20) = 10 minute spaces over the hour hand.

55 minute spaces are gained. in 60 min.

10 minute spaces will be gained in (60 x 10)/55 min. = 120/11min.

The hands will be in the same straight line but not together at 120/11 min.

Ans .

346/11 min. past 5

1. Explanation :

At 5 o'clock, the minute hand is 25 min. spaces behind the hour hand.

Case I. Minute hand is 3 min. spaces behind the hour hand.

In this case, the minute hand has to gain' (25 - 3) = 22 minute spaces. 55 min. are

gained in 60 min.

22 min. are gaineg in (60*22)/55min. = 24 min.

The hands will be 3 min. apart at 24 min. past 5.

Case II. Minute hand is 3 min. spaces ahead of the hour hand.

In this case, the minute hand has to gain (25 + 3) = 28 minute spaces. 55 min. are gained in 60 min.

28 min. are gained in (60 x 28_)/55=346/11

The hands will be 3 min. apart at 346/11 min. past 5.

Ans .

440/43 minutes

1. Explanation :

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes over the hour hand. 55 min. are gained in 60 min.

60 min are gained in (60/55) x 60 min =720/11 min.

But, they are together after 65 min.

Gain in 65 min =720/11-65 =5/11min.

Gain in 24 hours =(5/11 * (60*24)/65)min =440/43

The clock gains 440/43 minutes in 24 hours.

Ans .

Wednesday

1. Explanation :

Time from 8 a.m. on Sunday to 8 p.m. on following Sunday = 7 days 12 hours = 180 hours

The watch gains (5 + 29/5) min. or 54/5 min. in 180 hrs.

Now 54/5 min. are gained in 180 hrs.

5 min. are gained in (180 x 5/54 x 5) hrs. = 83 hrs 20 min. = 3 days 11 hrs 20 min.

Watch is correct 3 days 11 hrs 20 min. after 8 a.m. of Sunday.

It will be correct at 20 min. past 7 p.m. on Wednesday

Ans .

11 p.m

1. Explanation :

Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours.

Now 23 hrs 44 min. of this clock = 24 hours of correct clock.

356/15 hrs of this clock = 24 hours of correct clock.

89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.

= 90 hrs of correct clock.

So, the correct time is 11 p.m.

Ans .

48 min. past 12

1. Explanation :

Time from 8 a.m. on a day 1 p.m. on the following day = 29 hours.

24 hours 10 min. of this clock = 24 hours of the correct clock.

145 /6 hrs of this clock = 24 hrs of the correct clock

29 hrs of this clock = (24 x 6/145 x 29) hrs of the correct clock

= 28 hrs 48 min. of correct clock

The correct time is 28 hrs 48 min. after 8 a.m.

This is 48 min. past 12.

Ans .

Rs. 6480.,Rs. 4680,Rs. 5440

1. Explanation :

(i) Cost of Rs. 100 stock = Rs. 90

Cost of Rs. 7200 stock = Rs. (90/100 * 7200 ) = Rs. 6480.

(ii) Cost of Rs. 100 stock = Rs. (100+4)

Cost of Rs. 4500 stock = Rs. (104/100 * 4500 ) = Rs. 4680

(iii) Cost of Rs. 100 stock = Rs. (100-15)

Cost of Rs. 6400 stock = Rs. (85/100 * 6400 ) = Rs. 5440.

Ans .

Rs. (215/2),Rs. 3440

1. Explanation :

Cash required to purchase Rs. 100 stock = Rs (107+(1/2)) = Rs. (215/2).

Cash required to purchase Rs. 100 stock = Rs [(215/2)*(1/100)*3200] = Rs. 3440

Ans .

Rs. (383/4),Rs 2298

1. Explanation :

By selling Rs. 100 stock , cash realised = Rs. [(100-4)-(1/4)] = Rs. (383/4).

By selling Rs. 2400 stock, cash realised = Rs. [(383/4)*(1/100)*2400] = Rs 2298.

Ans .

Rs. 200

1. Explanation :

Income from Rs. 100 stock = Rs. 8.

Income from Rs. 2500 = Rs. [(8/1000*2500) =Rs. 200.

Ans .

Rs. 500

1. Explanation :

By investing Rs. 136, income obtained = Rs. 10.

By investing Rs. 6800, income obtained = Rs. [(10/136)*6800] = Rs. 500.

Ans .

7(1/2) % stock at 105 is better

1. Explanation :

Let the investment in each case be Rs. (105*94).

Case I : 7(1/2) 5 stock at 105:

On investing Rs. 105, income = Rs. (15/2).

On investing Rs. (105*94), income = Rs. [(15/2)*(1/105)*105*94] = Rs 705.

Case II : 6(1/2) % stock at 94:

On investing Rs. 94, income = Rs. (13/2).

On investing Rs. (105*94), income = Rs. [(13/2)*(1/94)*105*94] = Rs. 682.5.

Clearly, the income from 7(1/2) % stock at 105 is more.

Hence, the investment in 7(1/2) % stock at 105 is better.

cAns .

Rs. 912

1. Explanation :

Cost of 1 share = Rs. [(10-(3/4)) + (1/4)] = Rs. (19/2).

Cost of 96 shares = Rs. [(19/2)*96] = Rs. 912.

Ans .

Rs. 165, 6.2 %

1. Explanation :

Cost of 1 share = Rs. [25+5+1/4)] = Rs. (121/4).

Cost of 88 shares = Rs.[(121/4)*88] = Rs. 2662.

Face value of 88 shares = Rs. (88*25) = Rs. 2200.

Dividend on Rs. 100 = (15/2).

Dividend on Rs. 2200 = Rs. [(15/20*(1/100)*2200] = Rs. 165.

Income derived = Rs. 165.

Rate of interest on investment = [(165/2662)*100] = 6.2 %.

Ans .

Rs. 22.50

1. Explanation :

Suppose he buys each share for Rs. x.

Then, [25*(9/100)] = [x*(10/100)] or x = Rs. 22.50.

Cost of each share = Rs. 22.50.

Ans .

Rs. 4200.

1. Explanation :

S.P of Rs. 5000 stock = Rs. [(156/100)*5000] = Rs. 7800.

Income from this stock = Rs. [(12/100)*5000] = Rs. 600.

Let investment in * % stock be x and that in 9 % stock = (7800-x).

[x*(8/90)] + (7800-x) * (9/108) = (600+7)

(4x/45) + [(7800-x)/12] = 670 -->16x + 117000-15x = (670*180) --> x = 3600.

Money invested in 8 % stock at 90 = Rs. 3600.

Money invested in 9 % at 108 = Rs. (7800-3600) = Rs. 4200.

Ans .

870

1. Explanation :

We have, 30!/28! = 30x29x(28!)/28! = (30x29) = 870.

Ans .

205320,24

1. Explanation :

(i) $$_{60}^{3}\textrm{P}$$ = 60!/(60-3)! = 60!/57! = 60x59x58x(57!)/57! = (60x59x58) = 205320.

(ii) $$_{4}^{4}\textrm{P}$$ = 4! = (4x3x2x1) = 24.

Ans .

120,4950,1

1. Explanation :

(i) $$_{8}^{10}\textrm{C}$$ = $$\frac{10*9*8}{3!}$$ = 120.

(ii) $$_{98}^{100}\textrm{C}$$=$$_{(100-98)}^{100}\textrm{C}$$ = $$\frac{100*99}{2!}$$ = 4950.

(iii) $$_{50}^{50}\textrm{C}$$= 1. [$$_{n}^{n}\textrm{C}$$ = 1]

Ans .

120

1. Explanation :

The word BIHAR contains 5 different letters.

Required number of words = $$_{5}^{5}\textrm{P}$$ = 5! = (5x4x3x2x1) = 120.

Ans .

4320

1. Explanation :

Given word contains 8 different letters. When the vowels AUE are always together,

we may suppose them to form an entity, treated as one letter.

Then, the letters to be arranged are DGNTR (AUE).

Then 6 letters to be arranged in $$_{6}^{6}\textrm{P}$$ = 6! = 720 ways.

The vowels in the group (AUE) may be arranged in 3! = 6 ways.

Required number of words = (720x6) = 4320.

Ans .

72

1. Explanation :

The given word contains 5 different letters.

Taking the vowels EA together, we treat them as one letter.

Then, the letters to be arranged are XTR (EA).

These letters can be arranged in 4! = 24 ways.

The vowels EA may be arranged amongst themselves in 2! = 2 ways.

Number of words, each having vowels together = (24x2) = 48 ways.

Total number of words formed by using all the letters of the given words

= 5! = (5x4x3x2x1) = 120.

Number of words, each having vowels never together = (120-48) = 72.

Ans .

2160

1. Explanation :

In the given word, we treat the vowels IEO as one letter.

Thus, we have DRCTR (IEO).

This group has 6 letters of which R occurs 2 times and others are different.

Number of ways of arranging these letters = 6!/2! = 360.

Now 3 vowels can be arranged among themselves in 3! = 6 ways.

Required number of ways = (360x6) = 2160.

Ans .

1365

1. Explanation :

Required number of ways = $$_{11}^{15}\textrm{C}$$ = $$_{(15-11)}^{15}\textrm{C}$$ = $$_{4}^{11}\textrm{C}$$

= $$\frac{15*14*13*12}{4*3*2*1}$$ = 1365.

Ans .

200

1. Explanation :

(3 men out 6) and (2 ladies out of 5) are to be chosen.

Required number of ways = ($$_{3}^{6}\textrm{C}$$x$$_{2}^{5}\textrm{C}$$) = [$$\frac{6*5*4}{3*2*1}$$] x [$$\frac{5*4}{2*1}$$] = 200

Ans .

1/2

1. Explanation :

Here s={H,T} and E={H}.

P(E)=n(E)/n(S)=1/2

Ans .

3/4

1. Explanation :

Here S={HH,HT,TH,TT}

Let Ee=event of getting one head

E={TT,HT,TH}

P(E)=n(E)/n(S)=3/4

Ans .

1/3

1. Explanation :

Here S={1,2,3,4,5,6}

Let E be the event of getting the multiple of 3

then ,E={3,6}

P(E)=n(E)/n(S)=2/6=1/3

Ans .

5/12

1. Explanation :

Here n(S)=(6*6)=36

let E=event of getting a total more than 7

={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

P(E)=n(E)/n(S)=15/36=5/12.

Ans .

7/15

1. Explanation :

.let S be the sample space

Then n(S)=no of ways of drawing 2 balls out of (6+4)=($$_{2}^{10}\textrm{C}$$=(10*9)/(2*1)=45

Let E=event of getting both balls of same colour

Then n(E)=no of ways(2 balls out of six) or(2 balls out of 4)

=(($$_{2}^{6}\textrm{C}$$+($$_{2}^{4}\textrm{C}$$)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21

P(E)=n(E)/n(S)=21/45=7/15

Ans .

7/18

1. Explanation :

Clearly n(S)=6*6=36

Let E be the event that the sum of the numbers on the two faces is divided by

4 or 6.Then

E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}

n(E)=14.

Hence p(e)=n(e)/n(s)=14/36=7/18

Ans .

55/221

1. Explanation :

We have n(s)=52c2=(52*51)/(2*1)=1326.

Let A=event of getting both black cards

B=event of getting both queens

A$$\cap$$B=event of getting queen of black cards

n(A)= ($$_{2}^{26}\textrm{C}$$=(26*25)/(2*1)=325,

n(B)= ($$_{2}^{4}\textrm{C}$$=(4*3)/(2*1)=6 and

n(A$$\cap$$B)= ($$_{2}^{2}\textrm{C}$$=1

P(A)=n(A)/n(S)=325/1326;

P(B)=n(B)/n(S)=6/1326 and

P(A$$\cap$$B)=n(A$$\cap$$B)/n(S)=1/1326

P(A$$\cap$$B)=P(A)+P(B)-P(A$$\cap$$B)=(325+6-1/1326)=330/1326=55/221.

Ans .

Rs. 180

1. Explanation :

P.W=100 x Amount /[100 + (R x T)]

=Rs.100 x 930/100+ (8x3)

= (100x930)/124= Rs. 750,

T.D. = (Amount) - (P.W.) = Rs. (930 - 750) = Rs. 180.

Ans .

Rs. 6000

1. Explanation :

Let amount be Rs. x.

Then,x*R*T/100 + (R x T)

=T.D.=>x*12*(3/ 4)/[100+[12*3/4]]=540

x= 540x109 = Rs.6540

Amount - Rs. 6540. P.W. = Rs. (6540 - 540) - Rs. 6000..

Ans .

16 2/3%.

1. Explanation :

T.D. = Rs. 250 and S.I. = Rs. 375.

Sum due =S.I. xT.D./ S.I. -T.D.=375x250/375- 250=Rs.750.

Rate=[100*375/750*3]%=16 2/3%.

Ans .

Rs. 6800

1. Explanation :

Let the sum be Rs. x. Then,

T.D. = (x*25/2*1/2)/(100+(25/2*1/2))=x*25/4*4/425=x/17

S.I=x*25/2*1/2*1/100=x/16

x/16-x/17=25 =>17x-16x=25*16*17 =>x=6800

Hence, sum due = Rs. 6800..

Ans .

Rs. 3600

1. Explanation :

Let the sum be Rs. x. Then,

[x/2+(x/2*100)/100+(25/2*2)]-[(x*100)/(100+25/2*1]=40

=>x/2+2x/5-8x/9=40=>x=3600

Amount of the bill - Rs. 3600

Ans .

Rs. 5880

1. Explanation :

Face value of the bill = Rs. 6000.

Date on which the bill was drawn = July 14 at 5 months. Nominally due date =December 14.,Legally due date = December 17.

Date on which the bill was discounted = October 5. Unexpired time : Oct. Nov. Dec.

26 + 30 + 17 = 73 days =1/ 5Years

B.D. = S.I. on Rs. 6000 for 1/5 year= Rs. (6000 x 10 x1/5 x1/100)= Rs. 120.

T.D. = Rs.[(6000 x 10 x1/5)/(100+(10*1/5))]=Rs.(12000/102)=Rs. 117.64.

B.G. = (B.D.) - (T.D.) = Rs. (120 - 117.64) = Rs. 2.36. Money received by the holder of the bill = Rs. (6000 - 120)= Rs. 5880..

Ans .

Rs. 129

1. Explanation :

B.G. = S.I. on T.D.

= Rs.(120 x 15 x 1/2 x 1/100)= Rs. 9.

(B.D.) - (T.D.) = Rs. 9.

B.D. = Rs. (120 + 9) = Rs. 129.

Ans .

4 months

1. Explanation :

S.I. on Rs. 1800 = T.D. on Rs. 1872.

P.W. of Rs. 1872 is Rs. 1800.

Rs. 72 is S.I. on Rs. 1800 at 12%.

Time =[(100 x 72)/ (12x1800)]year=1/3year = 4 months.

Ans .

13 7/11%

1. Explanation :

Sum =[( B.D.*T.D.)/(B.D.-T.D.)] = Rs.[(120x110)/(120-110)]= Rs. 1320. Since B.D. is S.I. on sum due, so S.I. on Rs. 1320 for 8 months is Rs. 120. Rate =[(100 x120)/( 1320 x 2/3)%= 13 7/11%.

Ans .

Rs. 121

1. Explanation :

T.D. =$$\sqrt{(P.W.*B.G)}$$

B.G. =$$\frac{(T.D.)^2}{P.W}$$ = Rs.[(110x110)/ 1100]= Rs. 11.

B.D.= (T.D. + B.G.) = Rs. (110 + 11) = Rs. 121..

Ans .

Rs 15

1. Explanation :

Sum = [(B.D.xT.D.)/ (B.D.-T.D.)]= [(B.D.xT.D.)/B.G.]

T.D./B.G. = Sum/ B.D.=1650/165=10/1

Thus, if B.G. is Re 1, T.D. = Rs. 10.

If B.D.is Rs. ll, T.D.=Rs. 10.

If B.D. is Rs. 165, T.D. = Rs. [(10/11)xl65]=Rs.150

And, B.G. = Rs. (165 - 150) = Rs 15.

Ans .

13 1/3%

1. Explanation :

Let amount of the bill = Rs.100

Money deducted =Rs.10

Money received by the holder of the bill = Rs.100-10 = Rs.90

SI on Rs.90 for 10 months = Rs.10

Rate =[(100*10)/(90*10/12)%=13 1/3%

Ans .

$$60^{\circ}$$

1. Explanation :

Let AB be the pole and AC be its shadow.

Let angle of elevation,$$\angle$$ACB=$$\theta$$

Then, AB = 2$$\sqrt{3}$$m AC = 2 m.

Tan $$\theta$$=AB/AC = $$\sqrt[2]{\frac{3}{2}}$$=$$\sqrt{3}$$=>$$\theta$$

So, the angle of elevation is $$60^{\circ}$$

Ans .

9.5

1. Explanation :

Let AB be the wall and BC be the ladder.

Then, $$\angle$$ACB= $$60^{\circ}$$ and BC = 19 m.

Let AC = x metres

AC/BC = cos $$60^{\circ}$$ => x/19 = ½ =>x=19/2 = 9.5

Distance of the foot of the ladder from the wall = 9.5

Ans .

20.76 m

1. Explanation :

Let AB be the tower and C and D be the points of observation. Then,

AB/AD = tan $$60^{\circ}$$ = $$\sqrt{3}$$ => AD = AB/$$\sqrt{3}$$= h/$$\sqrt{3}$$

AB/AC = tan $$30^{\circ}$$ = 1/$$\sqrt{3}$$ AC=AB x$$\sqrt{3}$$ = h$$\sqrt{3}$$

CD = (AC-AD) = (h$$\sqrt{3}$$-h/$$\sqrt{3}$$)

(h$$\sqrt{3}$$-h/$$\sqrt{3}$$) = 24 => h=12$$\sqrt{3}$$=(12*1.73)=20.76

Hence, the height of the tower is 20.76 m.

Ans .

18 m

1. Explanation :

Let AB be the tree and AC be the river. Let C and D be the two positions of the man. Then,

$$\angle$$ACB=$$60^{\circ}$$,$$\angle$$ADB=$$30^{\circ}$$ and CD=36 m.

Let AB=h metres and AC=x metres.

AB/AD=tan $$30^{\circ}$$=1/$$\sqrt{3}$$ => h/(36+x)=1/$$\sqrt{3}$$ =>h=(36+x)/$$\sqrt{3}$$

AB/AC=tan $$60^{\circ}$$=$$\sqrt{3}$$ => h/x=$$\sqrt{3}$$

h=$$\sqrt{3}$$/x .....(2)

From (i) and (ii), we get:

(36+x)/$$\sqrt{3}$$x => x=18 m.

So, the breadth of the river = 18 m.

Ans .

5 minutes

1. Explanation :

Let AB be the tower and C and D be the two positions of the boat.

h/y=tan $$60^{\circ}$$ =$$\sqrt{3}$$ => y=h/$$\sqrt{3}$$

h/(x+y)=tan $$30^{\circ}$$=1/$$\sqrt{3}$$ => x+y=$$\sqrt{3}$$h

x=(x+y)-y = ($$\sqrt{3}$$ h-h/$$\sqrt{3}$$ )=2h/$$\sqrt{3}$$

Now, 2h/$$\sqrt{3}$$ is covered in 10 min.

h/$$\sqrt{3}$$ will be covered in (10*($$\sqrt{3}$$/2h)*(h/$$\sqrt{3}$$))=5 min

Hence, required time = 5 minutes.

Ans .

18 m

1. Explanation :

Let AB and CD be the two temples and AC be the river.

Then, AB = 54 m. Let AC = x metres and CD=h metres.

$$\angle$$ACB=$$60^{\circ}$$, $$\angle$$EDB=$$30^{\circ}$$

AB/AC=tan $$60^{\circ}$$=$$\sqrt{3}$$

AC=AB/$$\sqrt{3}$$=54/$$\sqrt{3}$$=(54/$$\sqrt{3}$$*$$\sqrt{3}$$*$$\sqrt{3}$$)=18m

DE=AC=18$$\sqrt{3}$$

BE/DE=tan $$30^{\circ}$$=1/$$\sqrt{3}$$

BE=(18$$\sqrt{3}$$*1/$$\sqrt{3}$$=18 m

CD=AE=AB-BE=(54-18) m = 36 m.

So, Width of the river = AC = 18$$\sqrt{3}$$m=18*1.73 m=31.14m

Height of the other temple = CD= 18 m

The following table gives the sales of batteries manufactured by a company

lit the years. Study the table and answer the questions that follow:
YEAR4AH7AH32AH35AH55AHT0TAL
1992 75 144 114 102 108 543
1993 90 126 102 84 426 528
1994 96 114 75 105 135 525
1995 105 90 150 90 75 510
1996 90 75 135 75 90 465
1997 105 60 165 45 120 495
1998 115 85 160 100 145 605

Ans .

32AH

1. Explanation :

The total sales (in thousands) of all the seven years for various batteries are:

For 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676

For 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694

For 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901

For 35 AH= 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601

For 55 AH= 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799.

Clearly, sales are maximum in case of 32AH batteries.

Ans .

39000

1. Explanation :

Required difference = [(84 - 45) x 1000] = 39000

Ans .

1997

1. Explanation :

The percentages of sales of 4AH batteries to the total sales in different years are:

For 1992 =(75*100/543)%=13.81%

For 1993=(90*100)/528%=17.05%

For 1994=(96*100/465)%=19.35%

For 1995=(105*100/495)%=20.59%

For 1996=(96*100/465)%=19.35%

For 1997=(105*100/495)%=21.21%

For 1998=(115*100/605)%=19.01%

Clearly, the percentage is maximum in 1997.

Ans .

35AH

1. Explanation :

From the table it is clear that the sales of 7AH batteries have

been decreasing continuously from 1992 to 1997.

Ans .

34%.

1. Explanation :

Required Percentage =(145-108)/108)*100 %=34.26%=34%.

Study the following table carefully and answer these questions:

NUMBER OF CANDIDATES APPEARED AND QUALIFIED IN A COMPETITIVE

EXAMINATION FROM DIFFERENT STATES OVER THE YEAR
 1997 1998 1999 2000 2001 App. Qual. App. Qual. App. Qual. App. Qual. App. Qual. M 5200 720 8500 980 7400 850 6800 775 9500 1125 N 7500 840 9200 1050 8450 920 9200 980 8800 1020 P 6400 780 8800 1020 7800 890 8750 1010 9750 1250 Q 8100 950 9500 1240 8700 980 9700 1200 8950 995 R 7800 870 7600 940 9800 1350 7600 945 7990 885

Ans .

12.35%

1. Explanation :

Required Percentage=(1020+1240) *100%=(2260*100)/18300%

(8800+9500)=12.35%

Ans .

11.84%.

1. Explanation :

Required Percentage= (850+920+890+980+1350) *100% (7400+8450+7800+8700+9800)

=(4990*100)/42150%

=11.84%.

Ans .

=11.15%

1. Explanation :

Required Percentage=(84+1050+920+980+1020)/(7500+9200+8450+9200+8800)*100%=(4810*100)/43150*) %

=11.15%

Ans .

8990

1. Explanation :

Required average =(8100+9500+8700+9700+8950)/5=44950/5=8990

Ans .

2001

1. Explanation :

The percentages of candidates qualified to candidates appeared from

State P during different years are:

For 1997= 780 * 100% =12.19% 6400

for 1998 = 1020*100 %=11.59% 8800

For 1999 = 890*100 %=11.41%; 7800

For 2000 = 1010* 100 % = 11.54%. 8 750

For 2001=1250*100 %= 12.82% 9750

Maximum percentage is for the year 2001..

Ans .

80%

1. Explanation :

Required Percentage =( 720 + 840 + 780 + 950 + 870) . x 100

980+1050+1020+1240+940=80%

The following table gives the percentage of marks obtained by seven students in six , different subjects in an examination. Study the table and

answer the questions based on it. The numbers in the brackets give the maximum marks in each subject.
StudentMaths Chemistry Physics Geography History Computer Science
Max Marks(160) (130) (120) (100) (60) (40)
Ayush 90 50 90 60 70 80
Aman 1008080408070
sajal906070709070
Rohit806580806060
Muskan 806585955090
Tanvi 707565854060

Ans .

449

1. Explanation :

Aggregate marks obtained by Sajal

= [(90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) +(90% of 60) + (70% of 40)] = 135 + 78 + 84 + 70 + 54 + 28 = 449.

Ans .

60%.

1. Explanation :

Aggregate marks obtained by Tarun .

= [(65% of 150) + (35% of 130) + (50% of 120) + (77% of 100) + (80% of60) + (80% of 40)] = 97.5 + 45.5 + 60 + 77 + 48 + 32 = 360.

Total maximum marks (of all the six subjects)

= (150 + 130 + 120 + 100 + 60 + 40) = 600.

Overall percentage of Tarun = 360 x 100 % = 60%.

Ans .

89.14

1. Explanation :

Average marks obtained in Physics by all the seven students

= 1/7 [(90% of 120) + (80% of 120) + (70% of 120) + (80% of 120)+ (85% of 120) + (65% of 120) + (50% of 120)]

= 1/7 [(90 + 80 + 70 +80 + 85 + 65 + 50)% of 120]

=1/7 [520% of 120] = 89.14.

Ans .

2

1. Explanation :

From the table it is clear that Sajal and Rohit have 60% or more

marks in each of the six subjects.

Ans .

Maths

1. Explanation :

We shall find the overall percentage (for all the seven students) with respect to each subject.The overall percentage for any subject is equal to the average of percentages obtained by all the seven students since the maximum marks for any subject is the same for all the students.Therefore, overall percentage for:

(i) Maths = [1/7(90+100+90+80+80+70+65)]%= [1/7(575)]% = 82.14%.

(ii) Chemistry = [1/7(50 + 80 + 60 + 65 + 65 + 75 + 35)]%= [1/7(430)]% = 61.43%. .

(iii) Physics = [1/7(90 + 80 + 70 + 80 + 85 + 65 + 50)]%=[1/7 (520)]% = 74.29%.

(iv) Geography = [1/7(60 + 40 + 70 + 80 + 95 + 85 + 77)]%= [1/7 (507)} = 72.43%.

(v) History = [1/7 (70 + 80 + 90+ 60 + 50 + 40 + 80)]%=1/7 [(470)]% = 67.14%.

(vi) Computer Science = [1/7 (80 + 70 + 70 + 60 + 90 + 60 + 80)]%= [1/7 (510)]% = 72.86%.

Clearly; this. percentage is highest for Maths.

Study the following table carefully and answer tbe questions given below:(Bank P.O. 2001) CLASSIFICATION OF 100 STUDENTS BASED ON THE MARKS OBTAINED BY THEM IN PHYSICS AND CHEMISTRY IN AN EXAMINATION
(Marks out Of 50) Subject40 and above30 and above20 and above10 and above0 and above
physics 9 32 80 92 100
chemistry 4 21 66 81100
(aggregate Average) 7 27 73 87 100

Ans .

27

1. Explanation :

We have 40% of 50 =(40/100 x 50)= 20.

Required number = Number of students scoring less than 20 marks in aggregate

= 100 - number of students scoring 20 and above marks in aggregate = 100 - 73 = 27.

Ans .

32

1. Explanation :

We have 60% of 50 =(60 /100x 50) = 30.

Required number = Number of students scoring 30 and above mark in Physics = 32

Ans .

6

1. Explanation :

Required difference = (Number of students scoring 30 and

above in mark in Chemistry) (Number of students scoring 30 and above marks in aggregate) = 27 - 21 = 6.

Ans .

29%

1. Explanation :

Number of students getting at least 60% marks in Chemistry

= Number of students getting 30 and above marks in Chemistry = 21.

Number of students getting at least 40% marks in aggregate

= Number of students getting 20 and above marks in aggregate = 73.

Required Percentage = (21/73x 100)% = 28.77% ≈29%.

Ans .

20-30

1. Explanation :

Since 66 students get 20 and above marks in Chemistry and out of

these 21 students get 30 and above marks, therefore to select top 35 students in Chemistry, the qualifying marks

should lie in the range 20-30.

The bar graph given below shows the foreign exchange reserves of a country(in million us$) from 1991-92 to 1998-99 .answer the questions basedon this graph. Ans . 1.5 1. Explanation : required ratio = 5040/3360 = 1.5 Ans . 100% 1. Explanation : foreign exchange reserve in 1997-98=5040 million us$

foreign exchange reserves in 1993-94=2520 million us$therefore increase=(5040-2520)=2520 million us$

therefore percentage increase=((2520/2520)*100)%=100%

Ans .

1992-93

1. Explanation :

there is an increase in foreign exchange reserves during the years 1992-

93,1994-951996-97,1997-98 as compared to previous year (as shown by bar graph)

the percentage increase in reserves during these years compared to previous year are

(1) for 1992-93 =[(3720-2640)/2640*100]% =40.91%

(2) for 1994-95=[(3360-2520)/2520*100]%=33.33%

(3) for 1996-97=[(4320-3120)/3120*100]%=38.46%

(4) for 1997-98=[(5040-4320)/4320*100]%=16.67%

Clearly, the percentage increase over previous year is highest for 1992-93.

Ans .

125%

1. Explanation :

Average foreign exchange reserves over the given period

= [_x (2640 + 3720 + 2520 + 3360 + 3120 + 4320 + 5040 + 3120) ] million US $= 3480 million US$.

Foreign exchange reserves in 1996-97 = 4320 million US $. . . Required Percentage = x 100 % = 124.14% .. 125%. 3480 . Ans . 3:5 1. Explanation : Average foreign exchange reserves over the given period = 3480 million US$.

The country had reserves above 3480 million US $during the years 199293,1996-97 and 1997-98 i.e., for 3 years and below 3480 million US$ during the

years 1991-92, 1993-94, 1994-95, 1995-96 and 1998-99 i.e., for 5 years.

Hence, required ratio = 3 : 5.

bar-graph provided on next page gives the sales of books (inthousand numbers) from six branches of a publishing company during two

consecutive years 2000 and 2001. Answer the questions based on this bargraph:

Ans .

560

1. Explanation :

total sales of branches B1,B3 and B5 for both the years (in thousand

numbers)=(80+105)+(95+110)+(75+95)=560

Ans .

73.17%

1. Explanation :

required percentage=[(70+80)/(95+110)*100]%=(150/205*100)%=73.17%

Ans .

80

1. Explanation :

average sales of all the six branches (in thousand numbers ) for the year 2000=1/6*(80+75+95+85+75+70)=80

Ans .

7/9

1. Explanation :

required ratio=(75+65)/(85+95)=140/180=7/9

Ans .

87.5%

1. Explanation :

average sales(in thousand numbers of branches B1,B3,and B6 in 2000=

1/3*(80+95+70)=245/3

average sales(in thousand numbers of branches B1,B2,and B3 in

2001=1/3*(105+65+110)=280/3

therefore required percentage=[((245/3)/(280/3))*100]%=(245/280*100)%=87.5%

The bar graph provided below gives the data of the production of paper(in thousand tonnes) by three different companies x,y and z over the years .study

the graph and answer the questions that followproduction of paper(in laks tonnes) by three companys x,yand z over the years

Ans .

2,00,00,000 tons

1. Explanation :

required difference =[(45-25)*1,00,000]tons=20,00,000 tons.

Ans .

Rs. 912

1. Explanation :

average production of company x in the period 1998-

2000=[1/3*(25+50+40)]=(115/3) lakh tons average production of company y in the period 1998-2000

s[1/3*(35+40+50)]=(125/3) lakh tonstherefore req ratio=(115/3)/(125/3)=115/125=23/25

Ans .

60%c

1. Explanation :

percentage increase in the production y from 1996-1999=[(40- 25)/25*100]%=(15/25*100)%=60%

Ans .

x and z both

1. Explanation :

average production (in lakh tons)in five years for the three company’s are:

for company x=[1/5*(30+45+25+50+40)]=190/5=38

for company y=[1/5*(25+35+35+40+50)]=185/5=37

for company z=[1/5*(35+40+45+35+35)]=190/5=38

therefore the average production of maximum for both the company’s x and z

Ans .

1997

1. Explanation :

Percentage change (rise/fall)in the production of Company Y in comparison

to the previous year, for different years are:

For 1997 = [((32-25)/25)*100]% = 40%

For 1998 = [((35-35)/25)*100]% = 0%

For 1999 = [((40-35)/35)*100]% = 14.29%

For 2000 = [((50-40)/40)*100]% = 25%

Hence, the maximum percentage rise/fall in the production of company Y is for 1997.

Ans .

1996

1. Explanation :

The percentages of production of company z to the production of company

z for various years are:

For 1996 = ((35/25)*100)%=140%; For 1997 = ((40/35)*100)% = 114.29%

For 1998 = ((45/35)*100)%=128.57%; For 1999 = ((35/40)*100)%=87.5%

For 2000 = ((35/50)*100)%=70%

Clearly, this percentage is highest for 1996.

Ans .

1996

1. Explanation :

The percentage increase in the amount invested in raw-materials as

compared to the previous year, for different years are:

For 1996 = [((225-120)/120)*100]% = 87.5%

For 1997 = [((375-225)/225)*100]% = 66.67%

For 1998 = [((525-330)/330)*100]% = 59.09%

For 2000 there is a decrease.

Ans .

1997

1. Explanation :

The percentage change in the amount invested in raw-materials and in the

value of sales of finished goods for different years are:

in year 1996 Percentage change in amount invested in raw-materials[((225-120)/120)*100]% = 87.5%

Percentage change in value of sales of finished goods[((300-200)/200)*100]% = 50%

in year 1997 Percentage change in amount invested in raw-materials[((375-225)/225)*100]% = 66.7%

Percentage change in value of sales of finished goods[((500-300)/300)*100]% = 66.67%

in year 1998 Percentage change in amount invested in raw-materials[((525-330)/330)*100]% = -12%

Percentage change in value of sales of finished goods [((400-500)/500)*100]% = -20%

in year 1999 Percentage change in amount invested in raw-materials[((525-330)/330)*100]% = 59.09%

Percentage change in value of sales of finished goods[((600-400)/400)*100]% = 50%

in year 2000 Percentage change in amount invested in raw-materials[((420-525)/525)*100]% = -20%

Percentage change in value of sales of finished goods [((460-600)/600)*100]% = -23.33%

Thus the percentage difference is same during the year 1997

Ans .

Rs.77.51akhs

1. Explanation :

Required difference = Rs. [(1/6)*(200+300+500+400+600+460)- (1/6)*(120+225+375+330+525+420)] lakhs

= Rs. [(2460/6)-(1995/6)] lakhs = Rs.(410-332.5)lakhs = 77.5 lakhs.

Ans .

49%

1. Explanation :

Required percentage = [(600/(375+300+525))*100]% = 48.78% =>49%

Ans .

1997

1. Explanation :

The difference between the amount invested in raw-material and the value

of sales of finished goods for various years are :

For 1995 = Rs.(200-120)lakhs = Rs. 80 lakhs

For 1996 = Rs.(200-225)lakhs = Rs. 75 lakhs

For 1997 = Rs. (500-375)lakhs = Rs. 125 lakhs

For 1998 = Rs. (400-330)lakhs = Rs. 70 lakhs.

For 1999 = Rs. (600-525)lakhs = Rs. 75 lakhs

For 2000 = Rs. (460-420)lakhs = Rs. 40 lakhs.

Clearly, maximum difference was during 1997

Directions(questions 211 to 214) : study the following bar-graph and answer the questions given below.

Production of fertilizers by a Company (in 10000 tonnes) over the Years

Ans .

5

1. Explanation :

average production of each year=$$\frac{25+40+60+45+65+75+80}{8}$$=48.75=49

Ans .

1995 and 1999

1. Explanation :

average production of 1996 and 1997=$$\frac{40+60}{2}$$=50

average of 2000 and 2001=$$\frac{50+75}{2}$$=62.5

average of 1999 and 2000=$$\frac{65+50}{2}$$=57.5

average of 1998 and 2000=$$\frac{45+50}{2}$$=47.5

average of 1995 and 1999=$$\frac{25+75}{2}$$=50

Ans .

25%

1. Explanation :

increase=60-45=15

%increase=(15/60)*100=25%

Ans .

2001

1. Explanation :

the percentage increase in production as compared to the

previous year is maximum in 2001

The following pie-chart shows the sources of funds to be collected by the National

Highways Authority of India (NHAI) for its Phase II projects. Study the pie-chart and answer

the questions that follow.

Total funds to be arranged for Projects (Phase II) =Rs.57,600 crores.

Ans .

External Assistance

1. Explanation :

20% of the total funds to be arranged = Rs.(20% of 57600) crores

= Rs.11520 crores Rs.11486 crores.

Ans .

($$187.2^{\circ}$$

1. Explanation :

Central angle corresponding to Market Borrowing = $$[\frac{29952}{57600}*360]^{\circ}$$= ($$187.2^{\circ}$$

Ans .

1:6

1. Explanation :

Required ratio = $$\frac{4910}{29952}$$= $$\frac{1}{6.1}$$ =$$\frac{1}{6}$$

Ans .

6%

1. Explanation :

Shortage of funds arranged through External Assistance

=Rs.(11486-9695) crores =Rs. 1791 crores.

therefore, Increase required in Market Borrowings =Rs. 1791 crores.

Percentage increase required = $$\frac{1791}{29952}$$* 100 % = 5.98 % = 6%

Ans .

Rs. 5401 crores

1. Explanation :

Amount permitted = (Funds required from Toll for projects of Phase II ) +(10 % of these funds)

=Rs. 4910 crores + Rs. (10% of 4910) crores

=Rs. (4910 + 491) crores = Rs. 5401 crores.

The pie-chart provided below gives the distribution of land (in a village) under

various food crops. Study the pie-chart carefully and answer the questions that follow.

DISTRIBUTION OF AREAS (IN ACRES) UNDER VARIOUS FOOD CROPS

Ans .

Rice, Wheat and Barley

1. Explanation :

The total of the central angles corresponding to the three crops which cover 50% of the

total area ,should be 180.Now, the total of the central angles for the given combinations are:

(i) Wheat,Barley and jowar =$$72^{\circ}$$+ $$36^{\circ}$$+ $$18^{\circ}$$ =$$126^{\circ}$$

(ii) Rice,Wheat and Jowar = $$72^{\circ}$$+$$72^{\circ}$$+$$18^{\circ}$$=$$162^{\circ}$$

(iii) Rice,Wheat and Barley = $$72^{\circ}$$+$$72^{\circ}$$+$$36^{\circ}$$=$$180^{\circ}$$

(iv)Bajra,Maize and Rice = $$18^{\circ}$$+ $$45^{\circ}$$+$$72^{\circ}$$ = $$135^{\circ}$$

Clearly:(iii) is the required combination.

Ans .

6

1. Explanation :

The area under any of the food crops is proportional to the angle corresponding to that crop.

Let the area under the rice production be x million acres.

Then, 18:72 = 1.5:x =>x=(72*15/18)=6

Thus, the area under rice production be = 6 million acres.

Ans .

3:1

1. Explanation :

Let the total production of barley be T tones and let Z acres of land be put under barley production.

Then, the total production of wheat =(6T) tones.

Also,area under wheat production = (2Z) acres.

$$\frac{Area Under Wheat Production}{Area Under Barley Production}$$=$$\frac{72^{\circ}}{36^{\circ}}$$=2

And therefore,Area under wheat = 2*Area under Barley = (2Z)acres

Now, yield per acre for wheat = (6T/2Z) tones/acre = (3T/Z) tones/acre

And yield per acre for barley = (T/Z) tones/acre.

Required ratio = $$\frac{3T/Z}{T/Z}$$= 3:1.

Ans .

33%

1. Explanation :

Let Z acres of land be put under barley production.

$$\frac{Area Under Wheat Production}{Area Under Barley Production}$$=$$\frac{75^{\circ}}{36^{\circ}}$$=2

Area under rice production = 2 * area under barley production = (2Z) acres.

Now,if p tones be the yield per acre of barley then ,yield per acre of rice=(p+50% of p) tones =(3/2 p) tones.

Total production of rice = (yield per acre) * (area under production)= (3/2 p)*2Z=(3pZ) tones.

And,Total production of barley = (pz) tones.

Percentage production of barley to that rice = (pZ/3pZ *100)%= 33 1/3%

Ans .

$$76.8^{\circ}$$

1. Explanation :

Initially,let t be the total area under considerations.

The area under wheat production initially was =(72/360 * t)acres = (t/5)acres.

Now,if the total area under consideration be increased by 5%,

then the new value of the total area= (105/100 t) acres.

Also,if the area under wheat production be increased by 12%,

then the new value of area under wheat =|$$\frac{t}{5}$$ +(12% of $$\frac{t}{5}$$)| acres = (112t/500)acres.

Central angle corresponding to wheat in the pie-chart

= $$[\frac{Area Under Wheat (new)}{Total area (new)}*360]^{\circ}$$ = $$[\frac{(112t/500)}{(105t/100)}$$*360]^{\circ}\) =$$76.8^{\circ}$$

The following pie-charts show the distribution of students of graduate and post graduate levels in seven different institute-M,N,P,Q,R,S and T in a town.

DISTRIBUTION OF STUDENTS AT GRADUATE AND POST-GRADUATE LEVELS IN SEVEN INSTITUTES-M,N,P,Q,R,S AND T.

Ans .

(b) 8463

1. Explanation :

Students of institute M at graduate level = 17% of 27300 = 4641.

Students of institute S at graduate level = 14% of 27300 = 3822

Total number students at graduate level in institutes M and S = 4641+3822=8463.

Ans .

(c) 6669

1. Explanation :

Required number = (15% of 24700) + (12% of 24700) = 3705 + 2964 = 6669

Ans .

(d) 8372

1. Explanation :

Required number = (18% of 27300) + (14% of 24700) = 4914 + 3458 = 8372.

Ans .

(d) 19:14

1. Explanation :

Required ratio =$$\frac{21% of 24700}{14% of 27300}$$ = $$\frac{21 * 24700}{14 * 27300}$$ = $$\frac{19}{14}$$

Ans .

(d)19:13

1. Explanation :

Required ratio =$$\frac{21% of 24700}{14% of 27300}$$ = $$\frac{(21 * 24700)}{(13* 27300)}$$ = $$\frac{19}{13}$$

Study the following pie-chart and the table and the answer the questions based on them.

PROPORTION OF POPULATION OF SEVEN VILLAGES IN 1997

Ans .

22000.

1. Explanation :

Let the population of village X be x

Then,38% of x=12160 => x = $$\frac{12160 * 100}{38}$$ =3200

Now ,if s be the population village S,then

16:11 = 32000 : s => s= $$\frac{11 * 32000}{16}$$= 22000.

Ans .

23:11

1. Explanation :

Let N be the total population of all the seven villages.

Then ,population of village T below poverty line = 46% of (21% of N) and population of village Z

below poverty line = 42% of (11% of N)

Required ratio =$$\frac{46%of 21% of N}{42% of 11% of N}$$ = $$\frac{46*21}{42*11}$$ = $$\frac{23}{11}$$

Ans .

15600

1. Explanation :

Population of village R = 32000(given)

Let the population of village Y be y.

Then, 16:15 = 32000 : y => y = $$\frac{15 * 32000}{16}$$= 30000.

Ans .

12760

1. Explanation :

Population of village Y in 1997 = 30000(given) .

Let the population village V in 1997 be v.

Then, 15:10 = 30000:v => v = $$\frac{10 * 30000}{15}$$= 20000.

Now population of village V in 1998 = 20000 + (10% 0f 20000) = 20000.

Population of village V below poverty line in 1998 = 58% of 22000 = 12760.

Ans .

2:1

1. Explanation :

Let the total population of all the seven villages in 1997 be N.

Then,population of village R in 1997 = 16% of N = 16/ 100 N

And population of village Z in 1997 = 11% of N =11/100 N

Population of village R in 1999 = {16/100N+(10% of 16/100 N)}=1760/10000 N

and population of village Z in 1999 = {11/100 N-(5% of 11/100 N)} = 1045/10000 N.

Now,population of village R below poverty line for 1999 = 51% of(1760/10000 N)

And population of village Z below poverty line 1999 = 42% of (1045/10000 N)

Required ratio =$$\frac{51% of 1760/10000 N}{42% of 1045/10000 N}$$ = $$\frac{51 * 1760 }{42 * 1045}$$ = $$\frac{2}{1}$$

In a school the periodical examination are held every second month. In a session

during Apr. 2001 – Mar. 2002, a student of Class IX appeared for each of the periodical exams. The aggregate marks

obtained by him in each periodical exam are represented in the line-graph given below. Study the graph and answer

the questions based on it.(S.B.I.P.O 2003)

MARKS OBTAINED BY A STUDENT IN SIX PERIODICAL EXAMS HELD IN EVERY TWO MONTHS DURING THE YEAR IN THE SESSION 2001-02

Ans .

112.5%

1. Explanation :

Required percentage = [(405/360)*100] % = 112.5 %

Ans .

381

1. Explanation :

Average marks obtained in all the periodical exams.

= (1/6)*[360+370+385+400+404] = 380.83 = 381.

Ans .

78.75%

1. Explanation :

Required percentage = [(370+385)/(500+500) * 100] % = [(755/1000)*100]% =75.5%

Ans .

none

1. Explanation :

As is clear from graph, the total marks obtained in periodical exams, go on increasing.

Since, the maximum marks for all the periodical exams are same , it implies that the percentage of marks also goes on increasing. Thus,

in none of the periodical exams, there is a fall in percentage of marks compared to the previous exam.

Ans .

Oct. 01

1. Explanation :

Percentage increases in marks in various periodical exams compared to the previous exams are:

For Jun. 01 = [(365-360)/360 * 100 ] % = 1.39 %

For Aug. 01 = [(370-365)/365 * 100 ] % = 1.37 %

For Oct. 01 = [(385-370)/370 * 100 ] % = 4.05%

For Dec. 01 = [(400-385)/385 * 100 ] % = 3.90 %

For Feb. 02 = [(405-400)/400 * 100 ] % = 1.25 %

The following line- graph the ratio of the amounts of imports by a Company to the amount of exports from that Company over the period from 1995 to 2001. The questions given below are based on this graph. (S.B.I.P.O 2001)Ratio of value of Import to Export by a Company over the Years

Ans .

4

1. Explanation :

The exports are more than the imports implies that the ratio of value of imports to exports is less than 1.

Now, this ratio is less than 1 in the years 1995,1996,1997 and 2000. Thus, there are four such years.

Ans .

1997

1. Explanation :

The imports are minimum proportionate to the exports implies that the ratio of the

value of imports to exports has the minimum value.

Now, this ratio has a minimum value of 0.35 in 1997, i.e., the imports are minimum proportionate to the exports in 1997.

Ans .

b.Rs.320 crores

1. Explanation :

Ratio of imports to exports in the years 1996=0.85.

Let the exports in 1996=Rs.320 crores.

Then,272/x =0.85 implies x=272/.85 = 320..

Ans .

1. Explanation :

The graph gives only the ratio of imports to exports for different

years. To find the percentage increase in imports from 1997 to 1998, we require more details such as the

value of imports or exports during these years. Hence, the data is inadequate to answer this question.

Ans .

d.Rs 420 crores

1. Explanation :

The ratio of imports to exports for the years 1998 and 1999 are 1.25 and 1.40 respectively.

Let the exports in the year 1998 = Rs. x crores

Then, the exports in the year 1999=Rs(500-x) crores.

1.25=250/x implies x=250/1.25=200

Thus the exports in the year 1999=Rs. (500-200)crores=Rs.300 crores

Let the imports in the year 1999=Rs. y crores

Then, 1.4=y/300 implies y=(300*1.4)=420.

Imports in the year 1999=Rs.420 crores..

Study the following line-graph and answer the question based on it.Number of vehicle Manufactured by Two Companies over the Years

(Numbers in thousands)

Ans .

26000

1. Explanation :

Total production of Company X from 1997 to 2002

= 119000+99000+141000+78000+120000+159000 = 716000

and total production of Company Y from 1997 to 2002

=139000+120000+100000+128000+107000+148000=742000

Difference=742000-716000=26000.

Ans .

21000

1. Explanation :

Require difference = 128000-107000 = 21000.

Ans .

119333

1. Explanation :

Average number of vehicles manufactured by Company X

= (91/6)* (119000 + 99000 + 141000 + 78000 + 120000 + 159000) = 119333.

Ans .

2000

1. Explanation :

The difference between the production of Companies X and Y in various years are.

For 1997 = (139000 – 119000) = 20000;

For 1998 = (120000 – 99000) = 21000;

For 1999 = (141000 – 100000) = 41000;

For 2000 = (128000 – 78000) = 50000;

For 2001 = (120000 – 107000) = 13000;

For 2003 = (159000 – 148000) = 11000;

Clearly, maximum difference was in 2000

Ans .

164

1. Explanation :

Required percentage = [( 128000/78000)* 100] % = 164 %.

The following line-graph gives the percent profit earned by two Companies X and Y

during the period 1996 – 2001. Study the line – graph and answer the questions that are based on on it.

Percentage Profit Earned by Two Companies X and Y over the Given years % profit/ loss = [(Income – Expenditure) / Expenditure] * 100

Ans .

(b). Rs. 297 crores

1. Explanation :

Profit percent ofcompany Y in 1997=35.

Let the income of company Y in 1997 be Rs.x crores

Then,35 = $$\frac{x-220}{220}$$ X 100 =>x =297

Income of company Yin 1997 = Rs.297crores.

Ans .

16:15

1. Explanation :

Let the incomes of the twocompanies X and Yin 1999 be Rs.x and let the

Expenditures of companies X and Y in 1999 be E1 and E2 respectively

Then, for Company X we have:

50= $$\frac{x-E1}{E1}$$ x 100 => $$\frac{50}{100}$$ = $$\frac{x}{E1}$$ -1 => x = $$\frac{150}{100}$$ E1

Also, for the Company Y we have:

60 = $$\frac{x-E2}{E2}$$ *100 =>$$\frac{60}{100}$$ = $$\frac{x}{E2}$$ -1 =>x = $$\frac{160}{100}$$ E2

From (i) and (ii),we get

$$\frac{150}{100}$$ E1=$$\frac{160}{100}$$ E2 => $$\frac{E1}{E2}$$ = $$\frac{160}{150}$$= $$\frac{16}{15}$$(Required ratio).

Ans .

15:22

1. Explanation :

Let the incomes in 2000 of companies X and Y be 3x and 4x respectively.And let

the expenditure in 2000 of companies X and Y be E1 and E2 respectively.

Then, for company X we have:

65= $$\frac{3x-E1}{E1}$$ x 100 => $$\frac{65}{100}$$ = $$\frac{3x}{E1}$$-1 =>E1=3x *$$\frac{100}{165}$$

For company Y we have:

50 = $$\frac{4x-E2}{E2}$$ *100 => $$\frac{50}{100}$$ = $$\frac{4x}{E2}$$-1 =>E2 = 4x* $$\frac{100}{150}$$

From (i)and(ii) we get:

$$\frac{E1}{E2}$$ = $$\frac{3x*(100/165)}{4x*(100/150)}$$ =$$\frac{3* 150}{4*165}$$ =$$\frac{15}{22}$$ (Required ratio)

Ans .

Rs.102crores

1. Explanation :

Let the expenditures of each of the Companies X and Y in 1996 be

Rs.xcrores.And let the income of Company X in 1996 be Rs.zcrores so that the income of Company Y in 1996 =Rs.(342-z)crores.

Then,for company X we have:

40= $$\frac{z-x}{x}$$ *100 => $$\frac{40}{100}$$ =$$\frac{ z}{x}$$ -1 => x = $$\frac{100z}{140}$$

Also for company Y we have:

45= $$\frac{(342-z)-x}{x}$$ *100 => $$\frac{45}{100}$$= $$\frac{ (342-z)}{x}$$ -1 =>x = $$\frac{(342 –z)}{145}$$* 100

From(i)and (ii) we get:

$$\frac{100z}{140}$$ = $$\frac{(342-z)*100}{145}$$ =>z = 168

Substituting z=168 in (i),we get: x=120

Total expenditure of companies X and Y in 1996=2x=Rs.240crores.

Total income of companies X and Y in 1996=Rs.342 crores.

Total profit =Rs.(342-240)crores =Rs.102 crores

Ans .

Rs.465crores

1. Explanation :

Let the income of company X in 1998 be Rs.x crores.

Then,55= $$\frac{x-200}{200}$$ *100 => x = 310.

Expenditure of Company X in 2001= Income of company X in 1998 = Rs.310crores

Let the income of company X in 2001 be Rs.z crores

Then,50 = $$\frac{z-310}{310}$$ *100 =>z = 465.

Income of company X in 2001 = Rs.465 crores.