1.Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from 4th Feb., 2005 to 18th April, 2005.
- Rs.35.50
- Rs.37.50
- Rs.39.50
- Rs.41.50
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Solution:
Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.
P = Rs.3000 and R = 6.25%p.a = 25/4%p.a
S.I. = Rs.(3,000*(25/4)*(1/5)*(1/100))= Rs.37.50
2.
Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly.
- Rs. 804.32
- Rs. 814.32
- Rs. 824.32
- Rs. 834.32
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Solution:
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.
3. One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the area of the field.
- 120 m2
- 140 m2
- 160 m2
- 180 m2
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Solution:
Other side = ((17) 2- (15)2)(1/2) = (289- 225)(1/2) = (64)(1/2) = 8 m.
Area = (15 x 8) m2 = 120 m2.
4. 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.
- Rs. 1091
- Rs. 1171
- Rs. 1261
- Rs. 1351
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Solution:
ex:- , 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.
.
5.A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere interest rate is increased by 8%, it would amount to bow much ?
- Rs. 791
- Rs. 794
- Rs. 893
- Rs. 992
20
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Solution:
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.
6.Find a number such that when 15 is subtracted from 7 times the number, the Result is 10 more than twice the number.
- 5
- 10
- 15
- 25
21
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Solution:
Description :Let the number be x. Then,
7x - 15 = 2x + 10 => 5x = 25 =>x = 5.
Hence, the required number is 5.
8. 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. ?
- Rate = 13%
- Rate = 15%
- Rate = 17%
- Rate = 19%
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Solution:
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
9R2 / 5 = 405
R2 =((405 * 5 ) / 9) = 225
R = 15. Rate = 15%.
9.A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find rate of interest.
- 13%
- 11%
- 9%
- 7%
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Solution:
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%
10.The simple interest on a sum of money is 4/9 of the principal .Find the time, if rate percent and time both are numerically equal.
- 4 years 11 months
- 5 years 9 months
- 6 years 8 months
- 7 years 7 months
39
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Solution:
EX:- Let sum = Rs. x. Then, S.l. = Rs. 4x/9
Let rate = R% and time = R years.
Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3.
Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months.
11. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.
- 25 and 14
- 26 and 15
- 28 and 17
- 30 and 19
43
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View Answer
Solution:
EX:- Let the number be x and y. Then,
.
x - y = 11 ----(1) and 1/5 (x + y) = 9 => x + y = 45 ----(2)
.
Adding (1) and (2), we get: 2x = 56 or x = 28. Putting x = 28 in (1), we get: y = 17.
.
Hence' the numbers are 28 and 17.
.
12. 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.696 and rs.605
- rs.676 and rs.625
- rs.656 and rs.645
- rs.646 and rs.655
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Solution:
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.
.
13.The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit's age by 4 years, what is Nikita's age?
- 10 years
- 12 years
- 14 years
- 16 years
50
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Solution:
Let Ankit's age be x years. Then, Nikita's age = 240/x years.
2*(240 /x ) - x = 4
480 - x2 = 4x
x2 + 4x - 480 = 0
(x+24)(x-20) = 0
x = 20.
Hence, Nikita's age = 12 years.
14.One year ago, the ratio of Gaurav's and Sachin's age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Sachin ?
- 12 years
- 36 years
- 48 years
- 60 years
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Solution:
Ex:- Let Gaurav's and Sachin's ages one year ago be 6x and 7x years respectively.
Then, Gaurav's age 4 years hence = (6x + 1) + 4 = (6x + 5) years.
Sachin's age 4 years hence = (7x + 1) + 4 = (7x + 5) years.
(6x+5)/(7x + 5) = 7/8
8(6x+5) = 7(7x + 5)
48x + 40 = 49x + 35
x = 5.
Hence, Sachin's present age = (7x + 1) = 36 years.
15.The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.
- 33 years
- 36 years
- 39 years
- 42 years
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Solution:
Let the son's present age be x years.
Then, father's present age = (3x + 3) years
(3x + 3 + 3) = 2 (x + 3) + 10
3x + 6 = 2x + 16
x = 10.
Hence, father's present age = (3x + 3) = ((3*10) + 3) years = 33 years.
17. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the present age of Rajeev ?.
- 10 years
- 12 years
- 14 years
- 16 years
65
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Solution:
Let Rajeev's present age be x years. Then,
Rajeev's age after 15 years = (x + 15) years.
Rajeev's age 5 years back = (x - 5) years.
x + 15 = 5 (x - 5)
x + 15 = 5x - 25
4x = 40
x = 10.
Hence, Rajeev's present age = 10 years.
18. If the sum of two numbers is 42 and their product is 437, then find the absolute difference between the numbers.
- 2
- 4
- 6
- 8
70
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Solution:
Let the numbers be x and y. Then,
x + y = 42 and xy = 437
x- y = sqrt[(x + y)2 - 4xy] = sqrt[(42)2 - 4 x 437 ] = sqrt[1764 - 1748] = sqrt[16] = 4.
Required difference = 4.
19. A number is as much greater than 36 as is less than 86. Find the number.
- 61
- 63
- 65
- 67
73
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View Answer
Solution:
(Let the number be x. Then,
x - 36 = 86 - x
2x = 86 + 36 = 122
x = 61.Hence, the required number is 61.
20.In a stream running at 2kmph,a motar boat goes 6km upstream and back again to the starting point in 33 minutes.find the speed of the motarboat in still water.
- 18 kmph
- 20 kmph
- 22 kmph
- 24 kmph
79
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Solution:
EX:-, let the speed of the motarboat in still water be x kmph.then,
6/x+2 +6/x-2=33/60
11x2-240x-44=0
11x2-242x+2x-44=0
(x-22)(11x+2)=0
x=22.
