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Solutions PDF 1

The document contains a series of mathematical solutions to various problems, including calculations involving least common multiples (LCM), divisibility rules, and algebraic equations. Each solution is presented with a corresponding answer and detailed steps for clarity. The problems range from simple arithmetic to more complex algebraic manipulations.

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tarunchakravarthi04
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30 views5 pages

Solutions PDF 1

The document contains a series of mathematical solutions to various problems, including calculations involving least common multiples (LCM), divisibility rules, and algebraic equations. Each solution is presented with a corresponding answer and detailed steps for clarity. The problems range from simple arithmetic to more complex algebraic manipulations.

Uploaded by

tarunchakravarthi04
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Solutions

S1. Ans.(d)
1 1 1 1 1
+ + + +
35 63 99 143 195
1 1 1 1 1
= + + + +
5×7 9×7 9×11 11×13 13×15
1 1 1 1 1 1 1 1 1 1 1
= [ − + − + − + − + − ]
2 5 7 7 9 9 11 11 13 13 15
1 1 1 1
= [ − ]=
2 5 15 15

S2.Ans(c)
Sol: LCM of (8, 15, and 18) = 360
Minimum number added to make it perfect cube = 360 + 152 = 512
Sum of digit of number which is added = 1+ 5 + 2 = 8

S3.Ans(b)
Sol: LCM of (5, 8, 12 and 15) = 120
For the greatest 4 digit number = 120k + 4
put k = 83
= 120 × 83 + 4
= 9964

S4. Ans.(c)
Sol. 509xy0 divisible by 3 if sum of digits
5+9+𝑥+𝑦 14+𝑥+𝑦
Divisible by 3 ⟹ = __________ (1)
3 3
Divisible by 11 ⟹ 5 + 9 + y - x = 11 ⟹ x - y = 3 _____________ (1)
Now from (1) x + y = 7, x - y = 3
x = 5, y=2
The number is 509520
S5. Ans.(c)
Sol. 9digit number will be divisible by factor of 36, by 9 and 4.
For divisible by 4
Largest possible value of Y = 8
Now for divisible by 9
2+𝑥+2+1+2+3+7+8+4 29+𝑥
=
9 9
Possible value of x = 7
Now,
11x² - 5y² = 11 × 49 – 5 × 64
= 539 – 320
= 219

S6. Ans.(c)
Sol. L.C.M of (3, 7, 11) = 237
Let the maximum number divisible by 231 is 11799,

Maximum number divisible


= 11799 – 18
= 11781
x = 8, y = 1
Now, (x + y)
=8+1=9

S7. Ans.(c)
Sol.
17 1 1 1 1 1
= 60 ⟹ 9 ⟹ 1 ⟹ 1 = 1
60 3+17 3+ 8 3+ 1 3+ 1
17 1+ 1+ 9 1+
9 1
1+
8 8
On Comparing both
1 1
1 = 1
𝑎+ 1 3+ 1
𝑏+ 1 1+ 1
𝑐+8 1+8

a = 3, b = 1, c = 1
Now,
(a + b + c) = (3 + 1 + 1) = 5

S8. Ans.(c)
Sol. L.C.M of (3, 7 and 11) = 231
Let the largest five-digit number = 10399

P now, largest five-digit no.


= 10399 – 4 = 10395
a = 9, b = 5
Now,
(a + b)³ = a³ + b³ + 3ab (a + b)
= (9 + 5)³ = 14³ = 2744

S9. Ans.(d)
Sol. [168, 210, 264]
210 – 168 = 42 = 2 × 3 × 7
264 – 210 = 54 = 2 × 3³
264 – 168 = 96 = 25 × 3
HCF of 42, 54 and 96 be 2 × 3 = 6
Now,
Remainder when 168 is divided by 6 is 0
So, x = 6 and y = 0
𝑦 0
Then = =0
𝑥 6
S10. Ans.(d)
Sol.
Let the No. be
P = 21 × 1 + 4 = 25
Q = 21 × 1 + 9 = 30
R = 21 × 1 + 8 = 29
Now,
(9𝑃−3𝑄+5𝑅)
21
9×25−3×30+5×29
=
21
280
= = 7 Remainder
21
S11. Ans.(d)
Sol. √14 + 6 √5 = √9 + 5 + 2.3 √5
2
= √(3)2 + (√5) + 2.3√5
2
= √(3 + √5)
= 3 + √5

S12. Ans.(c)
Sol. Let the required number is x.
(7)−1 × x = (3)³
𝑥
= 27 ⟹ x = 189
7

S13. Ans.(c)
Sol. Let the number is x
(−39)−1 ÷ 𝑥 = (−13)−1
(−39)−1 1
(−13)−1
= 𝑥 ⟹ (−39) × (−13)
1
x=
3

S14. Ans.(d)
Sol.
3 11 𝑥+2 121
√( ) =
5 25
𝑥+2
11 3 11 2
( ) =( )
5 5
𝑥+2
=2⟹x+2=6
3
x=4

S15. Ans.(a)
Sol.
𝑥
1 2 1 1
( 2 ) =( )
(5) × (7)2 35
2𝑥
1 2 1 1
( ) =( )
35 35
x=1

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