On what sum of money will the compound interest for 2 years at 5% per annum amount to rupees 768 point 75?

On what sum of money will the compound interest for 2 years at 5% per annum amount to rupees 768 point 75?

On what sum of money will the compound interest for 2 years at 5% per annum amount to rupees 768 point 75?
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Question 8 Simple and Compound Interest Exercise 8.1

Answer:

First Case:

Principal (P1) = Rs 9600

Rate (R1) = 10%

Period = (T) = 3 years 6 months

= years = 7 / 2 years

Simple interest = (P × R × T) / 100

= (9600 × 10 × 7) / (100 × 2)

We get,

= Rs 3360

Second case:

Simple interest = Rs 3360 × 2

= Rs 6720

Rate (R) = 7.5% p.a. and

Period (T) = 2 years

Therefore,

Principal = (S.I × 100) / (R × T)

= (6720 × 100) / (7.5 × 2)

= Rs (6720 × 100 × 10) / (75 × 2)

= 6720000 / 150

We get,

= Rs 44800

Video transcript

hello everybody welcome to lido learning channel my name is rajna chaudhary and we have to solve the question which is a sum of money invested at 7.5 per annum simple interest for two years produce produces twice as much interest as 96 double zero in three years and six months as ten percent per annum a simple interest so uh the meaning of question is that a sum of money was invested and it is giving us twice interest twice as the interest as this sum is giving in three years so we have to find first of all the interest from the second case because if we will find the interest from the second case so uh the interest for the first case is twice of it so i'm writing k second first of all so case second so in case second principle is given to us 9600 rate is given 10 percent time is also given three years and six months so first of all we have to convert this three years and six months into years so uh 3 plus 6 upon 12 so after calculation it would be 1 by 2 that means 3 whole and 1 by 2 after further calculation we have 7 upon 2 years let's apply it in the formula so simple interest would be p into r into t upon 100 so because it was 7 upon 2 that's why we can write it like this way now two zeros are cancelled out here two can cancel out ten five times and then after further calculation we would have rupees 3 3 6 0 as the simple interest so this is the simple interest for case second now simple interest for case 1 is twice of this so let's find out the simple interest for case 1 so twice of 3 3 6 0 mean rupees 6 7 to 0. now this is the simple interest we know rate also that rate is 7.5 percent time is two years we can calculate principle so it would be like we know that simple interest is equal to prt upon 100 simple interest we have calculated 6720 is equal to principle we have to calculate rate is 7.2 7.5 i'm sorry and time is 2 years upon 100. we can transpose these uh 7.5 to 100 to the right hand side so equation would be like 6 7 2 0 multiplied by hundred upon seven point five multiply by two so two can cancel out uh this fifty times and then the it can be further calculated so i'm not wasting time and writing the answer as principle is 4 4 8 0 zero so i hope you understand the method see you in my next video don't forget to like share and subscribe the channel thank you for watching

On what sum of money will the compound interest for 2 years at 5% per annum amount to rupees 768 point 75?
On what sum of money will the compound interest for 2 years at 5% per annum amount to rupees 768 point 75?

Q.1

Find the simple interest on Rs.68,000 at 16 2/3% per annum for 9 months

8000

8500

7500

9000

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

Q.2

Find Simple Interest on Rs.3000 at 6 1/4% per annum for the period from

4th Feb,2005 to 18th April,2005.

37.50

45.00

35.50

38.00

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.

Q.3

A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

2625

1615

1630

1625

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.

Q.4

A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple interest

interest rate is increased by 8%, it would amount to bow mucb ?

950

992

192

882

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.

Q.5

Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years and at the rate of 14% p.a.

for the period beyond five years.1£ he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

12,050

11,500

12,000

11,000

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.

Q.6

A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find the sum and rate of interests.

208 ,13%

208 ,15%

308 ,13%

308 ,15%

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

Q.7

At what rate percent per annum will a sum of money double in 16 years.

7\(\frac{1}{5}\)% p.a

6\(\frac{1}{3}\)% p.a

7\(\frac{1}{4}\)% p.a

6\(\frac{1}{4}\)% p.a

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.

Q.8

The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal.

6\(\frac{1}{4}\)% p.a., 6 yrs 5 months

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

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

6\(\frac{1}{4}\)% p.a., 6 yrs 9 months

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.

Q.9

The simple interest on a certain sum of money for 2 l/2 years at 12% per annum is Rs. 40

less tban the simple interest on the same sum for 3 ½ years at 10% per annum.Find the sum.

800

1000

900

750

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.

Q.10

A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate,

it would have fetched Rs. 360 more.Find the sum.

6500

5000

6000

5900

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

Q.11 What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?

325

225

300

425

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.

Q.12

A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%.

If the total annual income is Rs. 106, find the money lent at each rate.

700,900

650,650

650,700

650,900

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.

Q.13

Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.

612

620

650

510

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.

Q.14

Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.

2000

2109

3109

3000

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.

Q.15

Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly

830

825.32

825

824.32

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.

Q.16

Find the compound interest on Rs. 16,000 at 20% per annum for 9 months,compounded quarterly

2000

2522

3522

3500

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

Q.17

If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200,

find the compound interest on the same sum for the same period at the same rate.

1261

1360

1260

1161

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.

Q.18 In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?

2 ys

3 yrs

4 yrs

2 yrs 5 months

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.

Q.19

If Rs. 600 amounts to Rs. 683.20 in two years compounded annually, find the rate of interest per annum.

8% p.a.

10% p.a.

5% p.a.

7% p.a.

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.

Q.20

. If the compound interest on a certain sum at 16 (2/3)% to 3 years is Rs.1270, find the simple interest

on the same sum at the same rate and f or the same period.

1080

1050

1110

2050

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.

Q.21

The difference between the compound interest and simple interest on a certain

sum at 10% per annum for 2 years is Rs. 631. Find the sum.

Rs.53,110.

Rs.63,100.

Rs.60,000.

Rs.73,100.

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.

Q.22

The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000

in 2 years was Rs. 405. What was the rate of interest p.c.p.a. ?

7%.

11%.

13%.

15%.

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%.

Q.23

Divide Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years,

the interest being compounded at 4% per annum.

Rs.676 and rs.665.

Rs.676 and Rs.625.

Rs.776 and Rs.725.

Rs.576 and Rs.625.

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.

Q.24

a certain sum amounts to rs.7350 in 2 years and to rs.8575 in 3 years.find the sum and rate percent.

Rs.5200

Rs.5000

Rs.4400

Rs.5400

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

Q.25

.a sum of money amounts to rs.6690 after 3 years and to rs.10,035 after

6 years on compound interest.find the sum.

Rs.4460.

Rs.4660.

Rs.5560.

Rs.3460.

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.

Q.26

a sum of money doubles itself at compound interest in 15 years.in how many years will it beco,e eight times?

55 years

25 years

35 years

45 years

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

Q.27

What annual payment will discharge a debt of Rs.7620 due in 3years at 16 2/3% per annum interest?

Rs.4430

Rs.3550

Rs.3430

Rs.5430

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.

Q.28

1.Evaluate:(1)\(log_3\)(27) (2)\(log_7\)(1/343)(3)\(log_100\)(0.01)

-3,-3,1s

-3,-3,-1

3,3,1

3,-3,-1

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

Q.29

(i) \(\log_{7} 1\)(ii)\(\log_{34} 34\) (iii)\(36^{\log_{6} 4} \)

0,0,16

0,0,0

4,2,16

3,0,4

Ans .

0,0,16

Reference https://www.physicsforums.com/threads/how-to-write-log-in-latex.817954/


  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.

Q.30

if \(\log_{\sqrt{8}} x\)=3 (1/3), find the value of x.

28

26

32

30

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.

Q.31

(i) \(\log_{5} 3\)* \(\log_{27} 25\)(ii) \(\log_{10} 27\) – \(\log_{27} 9\)

3/2,6/5

2/3,5/6

3/2,5/6

2/3,6/5

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

Q.32

Simplify :(log 75/16-2 log 5/9+log 32/243)

0

log 2

1

log 4

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

Q.33

6.Find the value of x which satisfies the relation

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

3/5

7/2

1

5/6

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

Q.34

.Simplify:[1/\(\log_{xy} (xyz)\) +1/\(\log_{yz} (xyz)\)+1/\(\log_{zx} (xyz)\)]

3

1

0

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

Q.35

If log 2=0.30103,find the value of log 50.

0.69897

1

1.69897

2.69897

Ans .

1.69897
  1. Explanation :

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

Q.36

If log 2=0.3010 and log 3=0.4771,find the values of:

i) log 25 ii)log 4.5

1.39,0.65

1.98,0.32

2,1

4,3

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

Q.37

If log 2=.30103,find the number of digits in \(2^{56}\).

18

15

19

17

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

Q.38

One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the area of the field.

130 \(m^2\)

110 \(m^2\)

120 \(m^2\)

100 \(m^2\)

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 \(m^2\)

Q.39

A lawn is in the form of a rectangle having its sides in the ratio 2: 3. The

area of the lawn is (1/6) hectares. Find the length and breadth of the lawn.

23/33,60

11/3,55

6/5,65

33/3,50

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.

Q.40

Find the cost of carpeting a room 13 m long and 9 m broad with a carpet

75 cm wide at the rate of Rs. 12.40 per square metre.

Rs. 1934.40.

Rs. 2934.40.

Rs. 2234.40.

Rs. 1834.40.

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.

Q.41

If the diagonal of a rectangle is 17 cm long and its perimeter is 46 cm,

find the area of the rectangle.

150 \(cm^2\)

110 \(cm^2\)

120 \(cm^2\)

100 \(cm^2\)

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 \(cm^2\) .

Q.42

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.

30 cm

10 cm

20 cm

15 cm

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.

Q.43

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 calculated from these measurements.

0.5%.

0.8%.

1.8%.

1.6%.

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%.

Q.44

A rectangular grassy plot 110 m. by 65 m has a gravel path 2.5 m wide all round it on

the inside. Find the cost of gravelling the path at 80 paise per sq. metre.

Rs. 715

Rs. 450

Rs. 680

Rs. 550

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

Q.45

The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third

square whose area is equal to the difference of the areas of the two squares.

24 cm

14 cm

34 cm

28 cm

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.

Q.46

A room 5m 55cm long and 3m 74 cm broad is to be paved with square tiles. Find the

least number of square tiles required to cover the floor.

150

166

160

176

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

Q.47

Find the area of a square, one of whose diagonals is 3.8 m long

8.52 \(m^2\)

7.22 \(m^2\)

6.43\(m^2\)

4.2 \(m^2\)

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 \(m^2\) .

Q.48

The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas

7 : 25.

4 : 35.

3: 25.

4 : 25.

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.

Q.49

If each side of a square is increased by 25%, find the percentage change in its area

55.15%

58.30%

56.25%

60.25%

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%.

Q.50

If the length of a certain rectangle is decreased by 4 cm and the width is increased by

3 cm, a square with the same area as the original rectangle would result. Find the perimeter

of the original rectang

60 cm

40 cm

50 cm

70 cm

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.

Q.51

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq. m

is Rs. 270 and the cost of papering the four walls at Rs. 10 per m2 is Rs. 1720. If a door and 2

windows occupy 8 sq. m, find the dimensions of the room.

l=9,b=8,h=7

l=9,b=6,h=7

l=6,b=6,h=6

l=9,b=6,h=6

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.

Q.52

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm

94 \(cm^2\).

84 \(cm^2\)..

74 \(cm^2\).

64 \(cm^2\).

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 \(cm^2\).

Q.53

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm.

30\(cm^2\)

20\(cm^2\)

40\(cm^2\)

50\(cm^2\)

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\(cm^2\) .

Q.54

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.

Base = 700 m and Altitude = 200 m

Base = 800 m and Altitude = 300 m

Base = 900 m and Altitude = 300 m

Base = 600 m and Altitude = 200 m

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.

Q.55

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is

32cm. Find the area of the triangle.

60 \(cm^2\)

70 \(cm^2\)

50 \(cm^2\)

80 \(cm^2\)

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 \(cm^2\)

Q.56

Find the length of the altitude of an equilateral triangle of side 3\(\sqrt{3}\) cm.

6.5 cm.

5.5 cm.

7.5 cm.

4.5 cm.

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.

Q.57

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights

is 3 : 4. Find the ratio of their bases.

17 : 9.

16 : 9.

11 : 3

13 : 7

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.

Q.58

The base of a parallelogram is twice its height. If the area of the parallelogram

is 72 sq. cm, find its height.

6 cm

9 cm

7 cm

4 cm

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.

Q.59

Find the area of a rhombus one side of which measures 20 cm and 01 diagonal 24 cm.

384 \(cm^2\)

344 \(cm^2\)

388 \(cm^2\)

376 \(cm^2\)

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 \(cm^2\)

Q.60

The difference between two parallel sides of a trapezium is 4 cm. perpendicular

distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

28 cm and 25 cm

25 cm and 20 cm

27 cm and 23 cm

30 cm and 28 cm

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

Q.61

Find the length of a rope by which a cow must be tethered in order tbat it

may be able to graze an area of 9856 sq. metres.

54

52.

58.

56.

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.

Q.62

The area of a circular field is 13.86 hectares. Find the cost of fencing it at

the rate of Rs. 4.40 per metre.

Rs. 5808

Rs. 5765

Rs. 5450

Rs. 5907

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.

Q.63

The diameter of the driving wheel of a bus is 140 em. How many revolution, per

minute must the wheel make in order to keep a speed of 66 kmph ?

150.

450.

350.

250.

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.

Q.64

A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.

13 m.

15 m.

14 m.

11 m.

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.

Q.65

The inner circumference of a circular race track, 14 m wide, is 440 m. Find

radius of the outer circle.

84 m

74 m

76 m

64 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.

Q.66

Two concentric circles form a ring. The inner and outer circumferences of ring are

(352/7) m and (518/7) m respectively. Find the width of the ring.

7 m

8 m

4 m

5 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.

Q.67

A sector of 120', cut out from a circle, has an area of (66/7) sq. cm. Find the

radius of the circle.

5 cm.

3 cm.

2 cm.

4 cm.

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.

Q.68

Find the ratio of the areas of the incircle and circumcircle of a square.

3 : 5.

1 : 3.

3 : 2.

1 : 2.

Ans .

1 : 2.


  1. Explanation :

    Radius of incircle = (x/2)

    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.

Q.69

If the radius of a circle is decreased by 50%, find the percentage decrease

in its area.

85%

75%

70%

80%

Ans .

75%


  1. Explanation :

    Let original radius = R. New radius =(50/100) R = (R/2)

    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%

Q.70

Find the volume and surface area of a cuboid 16 m long, 14 m broad and 7 m high.

768 \(cm^2\)

668 \(cm^2\)

868 \(cm^2\)

568 \(cm^2\)

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 \(cm^2\).

Q.71

Find the length of the longest pole that can be placed in a room 12 m long

8m broad and 9m high.

16 m.

17 m.

15 m.

13 m.

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.

Q.72

Tbe volume of a wall, 5 times as high as it is broad and 8 times as long as

it is high, is 12.8 cu. metres. Find the breadth of the wall.

50 cm

40 cm

30 cm

20 cm

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

Q.73

Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to

construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar?

45000.

55000.

50000.

35000.

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.

Q.74

Water flows into a tank 200 m x 160 m througb a rectangular pipe of

1.5m x 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2 metres?

94min

95min

93min

96min

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

Q.75

Tbe dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is

2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.

7.84 kg.

8.04 kg.

8.56 kg.

8.44 kg.

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.

Q.76

The diagonal of a cube is 63cm. Find its volume and surface area.

225 \(cm^2\)

116 \(cm^2\)

210 \(cm^2\)

216 \(cm^2\)

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 \(cm^2\)

Q.77

The surface area of a cube is 1734 sq. cm. Find its volume.

5913 \(cm^3\).

4913 \(cm^3\).

3913 \(cm^3\).

4900 \(cm^3\).

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 \(cm^3\).

Q.78

A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of

equal cubes. Find the least possible number of cubes.

36.

37

40.

39.

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.

Q.79

A cube of edge 15 cm is immersed completely in a rectangular vessel

containing water . If the dimensions of the base of vessel are 20 cm x 15 cm, find the rise in water level.

10.50 cm

11.25 cm

10.25 cm

11.76 cm

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.

Q.80

Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new

cube. Find the surface area of the cube so formed.

486 \(cm^2\)

586 \(cm^2\)

466 \(cm^2\)

476 \(cm^2\)

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 \(cm^2\).

Q.81

If each edge of a cube is increased by 50%, find the percentage increase in

Its surface area.

120%

115%

125%

135%

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%

Q.82

Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their

surface areas.

1:6

1:7

1:9

1:8

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.

Q.83

Find the volume , curved surface area and the total surface area of a cylinder

with diameter of base 7 cm and height 40 cm.

987 \(cm^2\)

857 \(cm^2\)

957 \(cm^2\)

897 \(cm^2\)

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 \(cm^2\).

Q.84

If the capacity of a cylindrical tank is 1848 m3 and the diameter of its base

is 14 m, then find the depth of the tank.

11 m

13 m

12 m

14 m

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

Q.85

2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm

diameter. Find the length of the wire in metres.

112 m.

110 m.

111 m.

122 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.

Q.86

How many iron rods, each of length 7 m and diameter 2 cm can be made

out of 0.88 cubic metre of iron?

450

300

400

350

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.

Q.87

The radii of two cylinders are in the ratio 3: 5 and their heights are in tbe

ratio of 2 : 3. Find the ratio of their curved surface areas

2.5

2.0

1.5

3.0

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

Q.88

If 1 cubic cm of cast iron weighs 21 gms, then find the eight of a cast iron

pipe of length 1 metre with a bore of 3 cm and in which thickness of the metal is 1cm.

28.3 kg.

26.4 kg.

23.8 kg.

25.14 kg.

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.

Q.89

Find the slant height, volume, curved surface area and the whole surface

area of a cone of radius 21 cm and height 28 cm.

45cm

30cm

25cm

35cm

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

Q.90

Find the length of canvas 1.25 m wide required to build a conical tent of

base radius 7 metres and height 24 metres

340 m.

460 m.

400 m.

440 m.

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.

Q.91

The heights of two right circular cones are in the ratio 1 : 2 and the

perimeters of their bases are in the ratio 3 : 4. Find the ratio of their volumes.

10 : 37

7 : 22

11 : 35

9 : 32

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.

Q.92

The radii of the bases of a cylinder and a cone are in the ratio of 3 : 4 and It

heights are in the ratio 2 : 3. Find the ratio of their volumes.

7 : 8.

8 : 5.

9 : 8.

7 : 4

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.

Q.93

A conical vessel, whose internal radius is 12 cm and height 50 cm, is full of

liquid. The contents are emptied into a cylindrical vessel with internal radius 10 cm.

Find the height to which the liquid rises in the cylindrical vessel

24 cm

29 cm

34 cm

26 cm

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

Q.94

Find the volume and surface area of a sphere of radius 10.5 cm.

4853 \(cm^{3}\),1378 \(cm^{2}\)

4851 \(cm^{3}\),1386 \(cm^{2}\)

4801 \(cm^{3}\),1360 \(cm^{2}\)

4785 \(cm^{3}\),1289 \(cm^{2}\)

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 \(cm^{2}\)

Q.95

If the radius of a sphere is increased by 50%, find the increase percent in

volume and the increase percent in the surface area

245.5%, 130%.

222%, 120%.

237.5%, 125%.

235%, 125%.

Ans .

237.5%, 125%.


  1. Explanation :

    Let original radius = R. Then, new radius = (150/100)R=(3R/2)

    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%.

Q.96

Find the number of lead balls, each 1 cm in diameter that can be

a sphere of diameter 12 cm.

1728.7

1776

1754

1747.6

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

Q.97

How many spherical bullets can be made out of a lead cylinder 28cm high and

with radius 6 cm, each bullet being 1.5 cm in diameter ?

1776

1734

1645

1792

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.

Q.98

A copper sphere of diameter 18cm is drawn into a wire of diameter 4 mm

Find the length of the wire.

343m

256m

243m

213m

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

Q.99

Two metallic right circular cones having their heights 4.1 cm and 4.3 cm and

the radii of their bases 2.1 cm each, have been melted together and recast into a

sphere. Find the diameter of the sphere.

3.5.cm

4.2.cm

3.4.cm

4.7.cm

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

Q.100

A Cone and a sphere have equal radii and equal volumes. Find the ratio of the

sphere of the diameter of the sphere to the height of the cone.

1:7

1:2

1:3

1:5

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.

Q.101

Find the volume , curved surface area and the total surface area of a

hemisphere of radius 10.5 cm.

1039.5 \(cm^{2}\)

1037.5 \(cm^{2}\)

1041.5 \(cm^{2}\)

1034.5 \(cm^{2}\)

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 \(cm^{2}\).

Q.102

Hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm

and height 4 cm. How many bottles will be needed to empty the bowl

45

51

54

63

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.

Q.103

A Cone,a hemisphere and a cylinder stand on equal bases and have the same

height.Find ratio of their volumes.

1:2:2

1:3:3

1:2:3

1:1:3

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

Q.104

In a km race, A beats B by 28 metres or 7 seconds. Find A's timeover the course.

5 min. 6 s

3 min. 5 s

4 min. 3 s

7 min. 3 s

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

Q.105

A runs 1 ¾ times as fast as B. if A gives B a start of 84 m, bow far must

winning post be so that A and B might reach it at the same time?

196 m

186 m

200 m

206 m

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

Q.106

A can run 1 km in 3 min. 10 sec. and B can cover the same distance in 3

min. 20 sec. By what distance can A beat B ?

64 m

48 m

50 m

68 m

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

Q.107

In a 100 m race, A runs at 8km per hour. If A gives B a start of 4 m and

still him by 15 seconds, what is the speed of B ?

5.13 km/hr

5.76 km/hr

4.76 km/hr

4.06 km/hr

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.

Q.108

A, Band C are three contestants in a km race. If A can give B a start of 40 m

and A can give C a start of 64m how many metre's start can B give C ?

806 m

906 m

936 m

986 m

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.

Q.109

In a game of 80 points; A can give B 5 points and C 15 points. Then how

many points B can give C in a game of 60 ?

7 points

8 points

9 points

5 points

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

Q.110

What was the day of the week on, 16th July, 1776?

Wednesday

Monday

Sunday

Tuesday

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'.

Q.111

What was the day of the week on 16th August, 1947?

Sunday

Saturday

Monday

Friday

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'.

Q.112

What was the day of the week on 16th April, 2000 ?

Sunday

Saturday

Friday

Monday

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'.

Q.113

On what dates of Jull.2004 did Monday fall?

6th, 13th, 20th and 27th

5th, 12th, 19th and 26th

4th, 11th, 18th and 25th

3th, 10th, 17th and 24th

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.

Q.114

Find the angle between the hour hand and the minute hand of a clock when 3.25

29°

23°

49°

47*1/2°

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°

Q.115

At what time between 2 and 3 o'clock will the hands of a clock be together?

110/13 min. past 8

12/11 min. past 28

120/11 min. past 2

12/19 min. past 20

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.

Q.116

At what time between 4 and 5 o'clock will the hands of a clock be at right angle?

53/11 min. past 14

4/11 min. past 40

40/41 min. past 7

40/11 min. past 4

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

Q.117

Find at what time between 8 and 9 o'clock will the hands of a clock

being the same straight line but not together.

120/11 min

113/11 min

123/13 min

110/13 min

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.

Q.119

At what time between 5 and 6 o'clock are the hands of a clock

3minapart?

36/11 min. past 9

346/11 min. past 5.

117/11 min. past 5.

316/13 min. past 5.

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.

Q.120

The minute hand of a clock overtakes the hour hand at intervals of 65

minutes of the correct time. How much a day does the clock gain or lose?

140/43 minutes

410/42 minutes

450/41 minutes

440/43 minutes

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.

Q.121

A watch which gains uniformly, is 6 min. slow at 8 o'clock in the

morning Sunday and it is 6 min. 48 sec. fast at 8 p.m. on following Sunday.

When was it correct?

Wednesday

Friday

Thursday

Tuesday

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

Q.122

A clock is set right at 6 a.m. The clock loses 16 minutes in 24 hours.

What will be the true time when the clock indicates 10 p.m. on 4th day?

1 a.m

12 p.m

11 p.m

2 a.m

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.

Q.123

A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will

be the true time when the clock indicates 1 p.m. on the following day?

36 min. past 11

25 min. past 8

48 min. past 12

36 min. past 18

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.

Q.124

Find the cost of:

(i) Rs. 7200, 8% stock at 90;

(ii) Rs. 4500, 8.5% stock at 4 premium;

(iii) Rs. 6400, 10% stock at 15 discount.

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

Rs. 6440.,Rs. 4660,Rs. 5440

Rs. 6560.,Rs. 4120,Rs. 5340

Rs. 6320.,Rs. 3680,Rs. 4440

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.

Q.125

Find the cash required to purchase Rs. 3200, 7(1/2) % stock at 107

(brokerage (1/2) %)

Rs. (115/2),Rs. 1140

Rs. (215/2),Rs. 3440

Rs. (215/2),Rs. 1440

Rs. (315/2),Rs. 4440

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

Q.126

Find the cash realised by selling Rs. 2440, 9.5% stock at 4 discount

(brokerage (1/4) %)

Rs. (324/4),Rs 2198

Rs. (383/4),Rs 2298

Rs. (313/4),Rs 1298

Rs. (333/7),Rs 2228

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.

Q.127

Find the annual income derived from Rs. 2500, 8% stock at 106.

Rs. 300

Rs. 200

Rs. 150

Rs. 250

Ans .

Rs. 200


  1. Explanation :

    Income from Rs. 100 stock = Rs. 8.

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

Q.128

Find the annual income derived by investing Rs. 6800 in 10% stock at 136.

Rs. 300

Rs. 600

Rs. 400

Rs. 500

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.

Q.129

Which is better investment? 7(1/2) % stock at 105 or 6(1/2) % at 94.

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

6(1/2) % at 94 is better

7(1/2) % stock at 105 and 6(1/2) % at 94 is better

none of them

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.

Q.130

Find the cost of 96 shares of Rs. 10 each at (3/4) discount, brokerage being

(1/4) per share.

Rs. 812

Rs. 912

Rs. 921

Rs. 902

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.

Q.131

Find the income derived from 88 shares of Rs. 25 each at 5 premium,

brokerage being (1/4) per share and the rate of dividend being 7(1/2) % per annum.

Also, find the rate of interest on the investment.

Rs. 165, 6.2 %

Rs. 155, 5.5 %

Rs. 175, 7.2%

Rs. 150, 5.6%

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.

    Investment made = 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 %.

Q.132

A man buys Rs. 25 shares in company which pays 9 % dividend. The

money invested is such that it gives 10 % on investment. At what price did he buy

the shares?

Rs. 32.50

Rs. 25.50

Rs. 22.50

Rs. 35.50

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.

Q.134

A man sells Rs.5000, 12 % stock at 156 and uinvests the proceeds parity in

8 % stock at 90 and 9 % stock at 108. He hereby increases his income by Rs. 70.

How much of the proceeds were invested in each stock?

Rs. 4500.

Rs. 3500.

Rs. 3800.

Rs. 4200.

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.

Q.135

Evaluate: 30!/28!

770

870

850

800

Ans .

870


  1. Explanation :

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

Q.136

Find the value of (i) \(_{60}^{3}\textrm{P}\) (ii) \(_{4}^{4}\textrm{P}\)

205320,24

205408,32

205439,30

205310,24

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.

Q.137

Find the vale of (i) \(_{8}^{10}\textrm{C}\) (ii)\(_{98}^{100}\textrm{C}\) (iii) \(_{50}^{50}\textrm{C}\)

110,4550,1

120,4950,1

120,5050,0

100,5250,0

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]

Q.138

How many words can be formed by using all letters of the word “BIHAR”

110

100

120

80

Ans .

120


  1. Explanation :

    The word BIHAR contains 5 different letters.

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

Q.139

How many words can be formed by using all letters of the word ‘DAUGHTER’ so

that the vowels always come together?

4320

4110

3370

2350

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.

Q.140

How many words can be formed from the letters of the word ‘EXTRA’ so that the

vowels are never together?

126

118

72

110

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.

Q.141

How many words can be formed from the letters of the word ‘DIRECTOR’

So that the vowels are always together?

2060

1180

2110

2160

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.

Q.142

In how many ways can a cricket eleven be chosen out of a batch of

15 players ?

1365

1330

1265

1225

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.

Q.143

In how many ways, a committee of 5 members can be selected from

6 men and 5 ladies, consisting of 3 men and 2 ladies?

100

210

200

110

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

Q.144

In a throw of a coin ,find the probability of getting a head

1/5

1/3

1/2

1

Ans .

1/2


  1. Explanation :

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

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

Q.145

Two unbiased coin are tossed .what is the probability of getting atmost

one head?

3/4

3/2

1/4

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

Q.146

An unbiased die is tossed .find the probability of getting a multiple of 3

1/2

1/5

1

1/3

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

Q.147

In a simultaneous throw of pair of dice .find the probability of getting

the total more than 7

5/13

11/12

5/11

5/12

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.

Q.148

A bag contains 6 white and 4 black balls .2 balls are drawn at random.

find the probability that they are of same colour.

7/11

7/15

2/15

3/11

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

Q.149

Two dice are thrown together .What is the probability that the sum of the

number on the two faces is divided by 4 or 6

5/13

7/11

7/18

2/3

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

Q.150

Two cards are drawn at random from a pack of 52 cards.what is the

probability that either both are black or both are queen?

5/11

55/21

2/11

55/221

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.

Q.151

Find the present worth of Rs. 930 due 3 years hence at

8% per annum. Also find the discount.

Rs. 180

Rs. 102

Rs. 192

Rs. 132

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.

Q.152

The true discount on a bill due 9 months hence at 12%

per annum is Rs. Find the amount of the bill and its present worth..

Rs. 5000

Rs. 4500

Rs. 6000

Rs.5500

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..

Q.153

The true discount on a certain sum of money due 3 years

hence is Rb. 250 and the simple interest on the same sum for the same time and

at the same rate is Rs. 375. Find the sumand the rate percent.

16 2/3%.

16 1/5%.

11 5/3%.

13 1/3%.

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%.

Q.154

The difference between the simple interest and true discount on a certain sum

of money for 6 months at 12% per annum is Rs. 25. Find the sum.

Rs.5500

Rs. 6200

Rs.5800

Rs. 6800

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..

Q.155

A bill falls due in 1 year. The creditor agrees to accept immediate payment of the half and to defer the payment of the

other half for 2 years. By this arrangement ins Rs. 40. What is the amount of the bill, if the money be worth 12% ?.

Rs. 3200

Rs. 3000

Rs. 3600

Rs. 3500

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

Q.156

A bill for Rs. 6000 is drawn on July 14 at 5 months. It is discounted on 5th

October at 10%. Find the banker's discount, true discount, banker's gain and the money that the holder of the bill receives

Rs. 5800

Rs. 5880

Rs. 4680

Rs. 4380

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..

Q.157

If the true discount on a certain sum due 6 months hence at 15% is Rs. 120,

what is the banker's discount on the same sum for the same time and at the same rate?

Rs. 129

Rs. 109

Rs. 125

Rs. 119

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.

Q.158

The banker's discount on Rs. 1800 at 12% per annum is equal to the true

discount on Rs. 1872 for the same time at the same rate. Find the time.

5 months

3 months

4 months

2 months

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.

Q.159

The banker's discount and the true discount on a sum of money due 8 months

hence are Rs. 120 and Rs. 110 respectively. Find the sum and the rate percent.

13 7/11%

13 %

11 13/11%

11 17/13%

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%.

Q.160

The present worth of a bill due sometime hence is Rs. 1100 and the true

discount on the bill is Rs. 110. Find the banker's discount and the banker's gain.

Rs. 111

Rs. 121

Rs. 110

Rs. 119

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..

Q.161

The banker's discount on Rs. 1650 due a certain time hence is Rs. 165. Find

the true discount and the banker's gain.

Rs 12

Rs 11

Rs 15

Rs 13

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.

Q.162

What rate percent does a man get for his money when in discounting a bill

due 10 months hence, he deducts 10% of the amount of the bill?

13 1/3%

11 1/5%

7 1/3%

12 1/3%

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%

Q.163

If the height of a pole is 2\(\sqrt{3}\)metres and the length of its shadow is 2 metres,

find the angle of elevation of the sun..

\(30^{\circ}\)

\(45^{\circ}\)

\(60^{\circ}\)

\(90^{\circ}\)

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 \(60^{\circ}\)

Q.164

A ladder leaning against a wall makes an angle of \(60^{\circ}\) with the ground. If the

length of the ladder is 19 m, find the distance of the foot of the ladder from the wall.

7.5

9.5

6.5

8.5

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

Q.165

The angle of elevation of the top of a tower at a point on the ground is \(30^{\circ}\). On

walking 24 m towards the tower, the angle of elevation becomes \(60^{\circ}\). Find the height of the tower.

29 m

22 m

25.6 m

20.76 m

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.

Q.166

A man standing on the bank of a river observes that the angle subtended by a

tree on the opposite bank is \(60^{\circ}\). When he retires 36 m from the bank, he finds the angle to be \(30^{\circ}\).

Find the breadth of the river.

12 m

18 m

15 m

20s 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.

    Then, AD=(36+x)metres......(1)

    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.

Q.167

A man on the top of a tower, standing on the seashore finds that a boat coming towards him takes 10 minutes for the angle

of depression to change from \(30^{\circ}\) to \(60^{\circ}\). Find the time taken by the boat to reach the shore from this position.

3 minutes

6 minutes

5 minutes

4 minutes

Ans .

5 minutes


  1. Explanation :

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

    Let AB=h, CD=x and AD=y.

    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.

Q.168

There are two temples, one on each bank of a river, just opposite to each

other. One temple is 54 m high. From the top of this temple, the angles of depression of the top and the foot of the other

temple are \(30^{\circ}\) and \(60^{\circ}\) respectively. Find the width of the river and the height of the other temple.

18 m

18 m

18 m

18 m

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

Q.169

The total sales of all the seven years is the maximum for which battery ?

4AH

7AH

32AH

55AH

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.

Q.170

What is the difference in the number of 35AH batteries sold in 1993 and 1997 ?

Q.171

The percentage of 4AH batteries sold to the total number of batteries sold

was maximum in the year:

Q.172

In the case of which battery there was a continuous decrease in sales from 1992 to 1997 ?

Q.173

What was the approximate percentage increase in the sales of 55AH

batteries in 1998 compared to that in 1992 ?.

NUMBER OF CANDIDATES APPEARED AND QUALIFIED IN A COMPETITIVE

EXAMINATION FROM DIFFERENT STATES OVER THE YEAR

19971998199920002001
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

Q.174

Combining the states P and Q, together in 1998, what is the percentage

of the candidates qualified to that of the canditates appeared?

(a) 10.87%

(b) 11.49%

(c) 12.35%

(d) 12.54%

Ans .

12.35%


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