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PROBLEMS ON ‘DIGITS / NUMBERS / FRACTIONS’
Ex-1: The sum of five consecutive positive integers is 55. The sum of the squares
of the extreme terms is
(1) 308
(2) 240
(3) 250
(4) 180
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Sol:
Let the five consecutive in integers be x, x + 1, x + 2, x + 3 and x
+4
⇒ 5x + 10 = 55
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⇒ (x) + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 55
⇒ 5x= 45 ⇒ x = 9
∴ The numbers are 9, 10, 11, 12 &13
Sum of squares of extremes = 92 + 132 = 81 + 169 = 250
Choice (3)
Ex-2: The sum of two numbers is 18. The difference of the numbers is 4. Find
the numbers.
(2) 12, 8
(3) 16, 2
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(1) 11, 7
(4) 8, 10
→ (i)
→ (ii)
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x + y = 18
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Sol: Let the two numbers be x and y.
x–y=4
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Add equations (i) and (ii), we get
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⇒ x+ y = 18
⇒x–y=4
2x = 22 ⇒ x = 11
Substitute x = 11 in equation (i),
⇒ (11) + y = 18 ⇒ y = 7
You can solve these types of questions by observation.
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Choice (1)
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Ex-3: If two is subtracted from the denominator of a fraction, the fraction
becomes 1/3. Instead, if two is added to the numerator and one is subtracted from
the denominator, the fraction becomes 1/2. Find the fraction.
(1) 6/17
(2) 7/18
(3) 4/15
(4) 3/11
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Sol:
N
D
N
1
=
D−2 3
From statement one
⇒ 3N –D = -2
From 2nd statement
→ (i)
N +2 1
=
D −1 2
⇒ 2N – D = -5 → (ii)
(i) – (ii), we get
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Let the fraction be
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⇒ 3N – D = -2
⇒ 2N – D = -5
- +
+
N=3
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Substitute N = 3 in equation (i)
⇒ 3 (3) – D = -2 ⇒ 9 –D = -2 ⇒ D = 11
N 3
∴ =
D 11
Choice (4)
Ex- 4: The sum of a two- digit number and the number formed by reversing the
digits is 55. Find the number, if one of the digits is one more than the other.
(1) 32
(2) 23
(3) 32 (or) 23 (4) None of these
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Sol:
Let the two-digit number be xy.
Its numerical value is 10x + y.
Number formed by reversing the digits is yx.
Its numerical value is 10y + x
⇒10x + y + 10y + x = 55 (given)
⇒ 11x + 11y = 55
→ (i)
⇒ 11(x + y) = 55 ⇒ x+ y = 5
As one digit is one more than the other
Let x = y + 1 (or) Let y = x + 1
From equation (i)
From equation (i)
⇒y+1+y=5
⇒x+x+1=5
⇒ 2y = 4
⇒ 2x = 4
⇒ x=2
⇒ x=3
∴ xy = 32
∴ xy = 23
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∴ The number can be either 32 (or) 23.
Choice (3)
Ex- 5: A two-digit number is such that twice the ten’s digit added to eleven times
(2) 86
(3) 73
(4) 54
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(1) 48
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the units digit is equal to the number itself. Find the number.
Sol: Let the two-digit number be xy
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Its numerical value is 10x + y
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2x + 11y = 10x + y (given)
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⇒ 8x = 10y ⇒
5
x 10 5
= = ⇒ ⇒ x : y =5 : 4
4
y 8 4
As 54 is the only number where the above condition is satisfied, with the
given ratio, the number must be 54 uniquely.
Choice (4)
Ex-6: In a three digit number, the middle digit equals the average of its extreme
digits. The sum of its digits is 9. How many possibilities can it take?
(1) 8
(2) 6
(3) 7
(4) 5
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Sol:
Let the number be xyz.
x+z
⇒ x + z = 2 y → (i)
2
y=
Substitute equation (i) in equation (ii)
2y + y = 9
⇒ 3y = 9
⇒y=3
∴x+z=6
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And x + y + z = 9 → (ii)
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Various possibilities for (x, z) are (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) and (6, 0)
∴ The number has 6 possibilities.
{x ≠ 0; so (0 ,6) is not possible}
Choice (2)
Ex-7: A three digit number is equal to 17 times the sum of the digits. If 198 is
added to the number, the digits get reversed; also the sum of the extreme digits of
the original number is less than the middle digit by unity. Find the sum of digits of
the original number.
(2) 8
(3) 7
(4) 10
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(1) 9
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Sol:
Let the 3-digit number be xyz
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Its numerical value is 100x+ 10y+ z
Number formed by reversing the digits is xyz
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Its numerical value is 100z + 10y + x
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100x + 10y + z = 17 (x+ y + z) (given)
⇒ 83x – 7y – 16z = 0 → (i)
Also
(100 x + 10y +z) + 198 = 100z + 10y + x
⇒ 99z - 99x = 198
⇒ z - x = 2 → (ii)
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∴ The sum of the extreme digits of the original number is less than the middle digit
by unity,
∴x+z=y-1
→ (iii)
Hence, Z = x + 2 & y = 2x + 3 [from (ii) and (iii)]
⇒ 83x – 7 (2x + 3) – 16 (x + 2) = 0
⇒ 83x – 14x – 21 – 16x – 32 = 0
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⇒ 53x = 53 ⇒ x= 1
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Substitute y and z values in equation (i)
∴ y = 2(1) + 3 = 5
∴ z = (1) + 2 = 3
∴ The sum of the digits of the original number would be x + y + z = 1 + 5 + 3 = 9
Choice (1)
Ex-8: A three –digit number is such that the sum of its digits is 17. The middle
(2) 962
(3) 562 (4) 764
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(1) 863
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digit of the number is 6. If 297 is subtracted from the number, the hundred’s digit
and the unit’s digit of the number are interchanged. Find the number.
Sol:
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Let the three digit number be x6y. (As middle digit is given)
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Its numerical value is 100x + 60 + y.
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Number formed by reversing its digits is y6x.
Its numerical value is 100y + 60 + x
⇒ x + 6 + y = 17 (given)
⇒ x + y = 11 → (i)
Also, 100x + 60 + y – 297 = 100y + 60 + x
⇒ 99x - 99y = 297
⇒ x - y = 3 → (ii)
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Add equations (i) and (ii), we get
x + y = 11
x–y=3
2x = 14 ⇒x = 7
NOTE: This problem can be solved by observation.
Choice (4)
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From equation (i),
⇒ (7) + y = 11 ⇒ y = 4. ∴ The number is 764.
(1) 1870 (2) 1770
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Ex-9: Manish was asked to find 5/6 times a number and he instead multiplied it
by 6/5. As a result, he got an answer which was more than the correct answer by
649. What was the number?
(3) 1860
Sol:
Let the number be x.
6
5
⇒ x − x = 649
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6
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11x
= 649 ⇒ x = 59 × 30 ⇒ x = 1770
30
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⇒
(4) 1760
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∴ The number was 1770.
Choice (2)
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Ex-10: A two-digit number is formed by either subtracting 17 from nine times
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the sum of the digits or by adding 21 to 13 times the difference of the digits. Find
the number.
(1) 37
(2) 73
(3) 71
(4) cannot be determined
Sol:
Let the two-digit number be xy.
Its numerical value is 10x+ y
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⇒ 10x + y = 9 (x + y) – 17
⇒ x – 8y = -17
→ (i)
Also, 10x + y = 13 (x - y) + 21
(Here we taken as only (x-y ), because (y-x) gives factional values.)
→ (ii)
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⇒ 3x - 14y = -21
(i) × 3 – (ii) × 1, we get
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(i) × 3 ⇒ 3x - 24y = - 51
(ii) ×1 ⇒ 3x - 14y = - 21
- +
+
-10y = -30
⇒ y=3
Substitute y value in equation (i),
⇒ x – 8(3) = - 17 ⇒ x = -17 + 24 ⇒ x = 7
∴ The required number is 73.
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Choice (2)
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