1) The value of ¼ + 1/4.3 + 1/4.32 + 1/4.33 correct to four places of decimals is
Sol: ¼ + 1/12 + 1/36 + 1/108 = 27+9+3+1/108
-- > 40/108 -- > 10/27
= 0.3704
2) A boy was asked to find the value of 7/12 of a sum of money. Instead of multiplying the sum by 7/12 he divided it by 7/12 and thus his answer exceeded the correct answer By Rs.95. Find the correct?
Sol: Let sum = Rs. K
: . 12/7 k – 7k/12 = 95
-- > 144k – 49k/84 = 95 -- > k = 84
: . 7/12 k -- > 7/12 * 84 = Rs. 49
3) Simplify 3.7*3.7+2.3*2.3+2*3.7*2.3 / 4.6*4.6-3.4*3.4
Sol: -- > (3.7 + 2.3)2 / (4.6 + 3.4) (4.6 – 3.4)
-- > 36/8*1.2 -- > 36/9.6 = 360/96 = 3 ¾
4) If 2/3rd of a number is subtracted from 7/3 of the number, the result is 2 more than the number itself. Find the number.
Sol: Let the number be k
: . 7/3 k – 2/3 k = k+2 -- > 5/3 k = k+2
-- > 2/3 k = 2 -- > = 3
5) Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the Sum of the third and 52. The smallest number is?
Sol: Let the numbers be x, y, z
: . x/3 = y/4 = z/5 = k
: . x = 3k, y = 4k, z = 5k
: . 3k + 5k= 4k + 52
-- > k = 13
: . Smallest number = 39
6) The least number of five digits which is exactly divisible by 12, 15 and 18 is
1) 10010
2) 10015
3) 10020
4) 10080
Sol: Least number of 5 digits is 10000.
L.C.M of 12, 15, 18 is 180
On dividing 10000 by 180, the remainder is 100.
: . Required number = 10000 + (180 – 100) = 10080
7) Of the three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number is?
1) 24
2) 36
3) 72
4) 108
Sol: Let the 3rd number be x, Then, second number = 3x
: . First number = 3x/2
: . x+3x + 3x/2 = (44*3)
-- > 11x/2 = 132 -- > x = 24
Largest number -- > 3x
-- > 3*24 = 72
8) If a/2 = b/3 = c/5, the value of a+b+c/c is
1) 2
2) 5
3) ½
4) 1/5
Sol: a/2 = b/3 = c/5 = k
-- > a = 2k, b = 3k, c = 5k
: . a+b+c/c = 2k+3k+5k/5k = 10k/5k = 2
9) If 1/3rdof a number subtracted from ½ of that number, then the difference is 10more than 1/7th of the same number. How much is that number?
Sol: Let the number = x
: . x/2 – x/3 = x/7+ 10
-- > 3x-2x/6 = x/7+10
-- > x/6 – x/7 = 10
-- >7x-6x/42 = 10 -- > x = 420
10) Simplify: 2 ½ of ¾ * ½ /3/2 + ½ / 3/2 (2/3 – ½ of 2/3) to get
1) 1 5/8
2) 2 5/8
3) 1 8/11
4) 3 5/8
Sol: 1) 2 ½ of ¾ * ½ / 3/2 + ½ / 3/2 (2/3 – ½ of 2/3)
-- > (5/2 * ¾) * ½ * 2/3 + ½ * 2/3 * (2/3 – ½ * 2/3)
-- > 15/8 * 1/3 + 1/3 * [ ( 2/3 – 1/3 )]
-- > 15/8 * 1/3 + 1/3 * [(2/3- 1/3) ]
-- > 5/8 + 1/3 / 1/3
-- > 5/8 + 1 = 1 5/8
11) Find the smallest number which when divided by 6,10 and 15 respectively leaves 5 as Remainder in each case?
Sol: L.C.M of 6,10,15 -- > 30
Required number =30 + 5 = 35
12) If a = 16 and b = then a2+b2+ab/a3-b3 will be
1) 11
2) 1/11
3) 121
4) Non
Sol: 1) 162+52+80 / 163-53 = 256+25+80/4096-125
= 361/3971 = 1/11
13) Simplify 2/3 / 4/9 of 7 ½ + 999 494/495 * 99
1) 99999
2) 10000
3) 99000
4) 11111
Sol: 2/3 / 4/9 of 15/2 + (999 +494/495) * 99
= 2/3 / 10/3 + 999*99 + 494/495 * 99
= 2/3 * 3/10 + (1000 – 1) 99 + 494/495 * 99
= 1/5 + 99000 – 99 + 494/5
= 494/5 + 99000 – 99 = 99000
14) If we multiply a fraction by itself and divide the product by its reciprocal; the fraction thus obtained is 15 5/8. The original fraction is?
Sol: Let the original fraction be x/y
X/y *x/y/y/x = 15 5/8
-- > x/y * x/y * x/y =125/8
-- > x3/y3 = 125/8
> x/y = 3√125/8 = 5/2 = 2 1/2
15) Find the H.C.F. of 852, 1065 and 1491.
Sol: 852) 1065(1
852
213) 852 (4
852
0
213) 1491(7
1491
0
Hence 213 is H.C.F of given numbers.
16) Find the greatest number which divides 3460 and 9380 leaving as remainder 9 and 13 respectively?
Sol: Since on dividing 3460 remainder 9 is left, the required number must divide (3460-9) = 3451 exactly. Similarly it must divide 9380 – 13 = 9367 exactly. Hence H.C.F of 9367 and 3451 will be required number.
Submitted by M.U.Subbaiah
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