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The difference of compound interest on Rs.800 for 1 year at 20% per annum when compounded half-yearly and quarterly is :
Solution:

Principal = Rs. (100×804×2)(100×804×2) = Rs.1000
C.I = Rs. [{1000×(1+4100)2−1000}][{1000×(1+4100)2−1000}] = Rs.81.60
1061.The difference between simple interest and the compound interest on Rs.600 for 1 year at 10% per annum, reckoned half-yearly is:
Rs.2.50
Rs.4.40
Rs.6.60
Nil
Explanation:

C.I when reckoned half-yearly
= Rs. [800×(1+10100)4−800][800×(1+10100)4−800] = Rs. 172.40
Difference = Rs.(172.40-168) Rs.4.40
1062.The compound interest of Rs.20480 at 614614% per annum for 2 years 73 days is:
Rs.6.60
Rs.4.40
Rs.1.50
Rs.3000
Explanation:

S.I = Rs. (600×10×1100)(600×10×1100) = Rs.60
C.I = Rs. [600×(1+5100)2−600][600×(1+5100)2−600] = Rs.61.50
Difference = Rs.(61.50-60) = Rs.1.50
1063.The compound interest on Rs.2800 for 112112years at 10% per annum is:
Rs.3131
Rs.2929
Rs.3636
Rs.441.35
Explanation:

73 days is 1/5 th of an year.
C.I = Rs.[20480×(1+254×100)2(1+15×254×100)]−20480[20480×(1+254×100)2(1+15×254×100)]−20480
= Rs. [(20480×1716×1716×8180)]−20480[(20480×1716×1716×8180)]−20480
= Rs.2929
1064.If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 2 years the amount to be paid is:
Rs.436.75
Rs.434
Rs.420
Rs.8082
Explanation:

Amount = Rs.[[2800×(1+10100)](1+5100)][[2800×(1+10100)](1+5100)]
= Rs.[2800×11100×2120][2800×11100×2120] = Rs.3234
C.I = Rs.(3234-2800)= Rs.434
1065.Which one of the following is not a prime number?
31
61
71
91
Explanation:

91 is divisible by 7. So, it is not a prime number.
1066.(112x54)=?;
67000
70000
76500
77200
Explanation:

(112*54) = 112*(10/2)4
=112*104/24
=1120000/16
=70000
1067.It is being given that (232 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?
216+1
216-1
7x223
296+1
Explanation:

Let 232=x
Then, (232+1)=(x+1)
Let x+1 be completely divisible by natural number N, then
(296+1)=[(232)3+1]=(x3+1)
=(x+1)(x2-x+1), which is completely divisible by N, since (x+1) divisible by N
1068.What least number must be added to 1056, so that the sum is completely divisible by 23 ?
2
18
3
21
Explanation:

If the number 1056 is completely divisible by 23 means, remainder should come zero.
But if we divide 1056 by 23, the remainder is 2.
So if 2 is added to the 1056, we get remainder 0.
Therefore solution is 2
1069.1397 397 = ?
1951609
1981709
18362619
2031719
Explanation:

1397 x 1397 = (1397)2
= (1400 - 3)2
= (1400)2 + (3)2 - (2 x 1400 x 3)
= 1960000 + 9 - 8400
= 1960009 - 8400
= 1951609.
1070.How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336
4
5
6
7
Explanation:


By using your calculator you can calculate that the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.
1071.The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?
240
270
295
360
Explanation:


Let the smaller number be x. Then larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Smaller number = 270.
1072.If the number 517 * 324 is completely divisible by 3, then the smallest whole number in the place of * will be:
0
1
2
None of these
Explanation:

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4)
= (22 + x), which must be divisible by 3.
x = 2.
1073.The Probability that a number selected at random from the first 50 natural numbers is a composite number is:
21/25
17/25
4/25
8/25
Explanation:

The number of exhaustive events = ⁵⁰C₁ = 50.
We have 15 primes from 1 to 50.
Number of favourable cases are 34.
Required probability = 34/50 = 17/25.
1074.A coin tossed live times. What is the probability that there is at least one tail?
31/32
1/16
1 / 2
1/32
Explanation:

Let P(T) be the probability of getting least one tail when the coin is tossed five times.
= There is not even a single tail.
i.e. all the outcomes are heads.
= 1/32 ; P(T) = 1 - 1/32 = 31/32
1075.If a number is chosen at random from the set {1,2,3,……100}, then the probability that the chosen number is a perfect cube is?
1/25
½
4/13
1/10
Explanation:

We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.
1076.If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is?
1 /4
1 / 2
3 / 4
3 / 5
Explanation:

The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
1077.Three 6 faced dice are thrown together. The probability that all the three show the same number on them is?
1 / 216
1 / 36
5 / 9
5 / 12
Explanation:

It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36
1078.A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.
62/63
125/126
1/63
1/126
Explanation:

Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126
1079.A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good
6/63
2/63
125/126
1/63
Explanation:

Required probability = 1 - 1/126 = 125/126
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