SOLVED EXAMPLES
Ex. 1. Simplify : (i) 8888 +
888 + 88 + 8
(ii)
11992 - 7823 - 456
Sol.
i ) 8888 ii) 11992 - 7823 - 456 =
11992 - (7823 + 456)
888 = 11992 - 8279 = 3713-
88 7823 11992
+ 8
+ 456 - 8279
9872
8279 3713
Ex. 2, What value will replace the question mark in each of the
following equations ?
(i) ? - 1936248 = 1635773
(ii) 8597 - ? = 7429 - 4358
Sol.
(i) Let x - 1936248=1635773.Then,
x = 1635773 + 1936248=3572021.
(ii) Let 8597 - x = 7429 - 4358.
Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.
Ex. 3. What could be the
maximum value of Q in the following equation? 5P9 + 3R7 + 2Q8 = 1114
Sol. We may analyse the given equation
as shown : 1 2
Clearly, 2 + P + R + Q = ll.
5 P 9
So, the maximum value of Q can be
3 R 7
(11 - 2) i.e., 9 (when P = 0, R = 0); 2 Q 8
11 1 4
Ex. 4. Simplify : (i) 5793405 x 9999
(ii) 839478 x 625
Sol.
i)5793405x9999=5793405(10000-1)=57934050000-5793405=57928256595.b
ii) 839478 x 625 = 839478 x 54 = 8394780000 =
524673750.
16
Ex. 5. Evaluate : (i) 986 x 237 + 986 x
863 (ii) 983 x 207 - 983 x 107
Sol.
(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.
(ii) 983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300.
Ex. 6. Simplify : (i) 1605 x 1605
ii) 1398 x 1398
Sol.
i) 1605 x 1605 = (1605)2 = (1600 + 5)2 =
(1600)2 + (5)2 + 2 x 1600 x 5
= 2560000 + 25 + 16000 = 2576025.
(ii) 1398 x 1398 = (1398)2 = (1400 - 2)2=
(1400)2 + (2)2 - 2 x 1400 x 2
=1960000 + 4 - 5600 = 1954404.
Ex. 7. Evaluate : (313 x 313 + 287 x 287).
Sol.
(a2 + b2)
= 1/2 [(a + b)2 + (a- b)2]
(313)2 + (287)2 = 1/2 [(313 + 287)2
+ (313 - 287)2] = ½[(600)2 + (26)2]
= 1/2 (360000 + 676) = 180338.
Ex. 8. Which of the following are prime numbers ?
(i) 241 (ii)
337 (Hi) 391 (iv) 571
Sol.
(i) Clearly, 16 > Ö241. Prime numbers
less than 16 are 2, 3, 5, 7, 11, 13.
241 is not
divisible by any one of them.
241 is a prime
number.
(ii) Clearly, 19>Ö337. Prime numbers less
than 19 are 2, 3, 5, 7, 11,13,17.
337 is not
divisible by any one of them.
337 is a prime
number.
(iii) Clearly, 20 > Ö39l". Prime numbers
less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.
We find
that 391 is divisible by 17.
391 is not prime.
(iv) Clearly, 24 > Ö57T. Prime numbers less
than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.
571 is
not divisible by any one of them.
571 is a
prime number.
Ex. 9. Find the unit's
digit in the product (2467)163 x (341)72.
Sol. Clearly, unit's digit in the given
product = unit's digit in 7153 x 172.
Now, 74 gives unit digit
1.
7152 gives unit digit 1,
\ 7153 gives unit
digit (l x 7) = 7. Also, 172 gives unit digit 1.
Hence, unit's digit in
the product = (7 x 1) = 7.
Ex. 10. Find the unit's digit in (264)102 + (264)103
Sol. Required unit's digit = unit's
digit in (4)102 + (4)103.
Now, 42 gives
unit digit 6.
\(4)102 gives unit digit 6.
\(4)103 gives unit digit of the product (6 x 4) i.e., 4.
Hence, unit's digit in
(264)m + (264)103 = unit's digit in (6 + 4) = 0.
Ex. 11. Find the total number
of prime factors in the expression (4)11 x (7)5 x (11)2.
Sol. (4)11x (7)5 x
(11)2 = (2 x 2)11 x (7)5 x (11)2 =
211 x 211 x75x 112 = 222
x 75 x112
Total number of prime
factors = (22 + 5 + 2) = 29.
Ex.12. Simplify : (i) 896 x
896 - 204 x 204
(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114
(iii) 81 X 81 + 68 X 68-2 x 81 X 68.
Sol.
(i) Given exp =
(896)2 - (204)2 = (896 + 204) (896 - 204) = 1100 x
692 = 761200.
(ii) Given exp = (387)2+
(114)2+ (2 x 387x 114)
= a2
+ b2 + 2ab, where a =
387,b=114
= (a+b)2 = (387 + 114 )2
= (501)2 = 251001.
(iii) Given exp = (81)2 + (68)2 – 2x 81 x 68 =
a2 + b2 – 2ab,Where a =81,b=68
=
(a-b)2 = (81 –68)2 = (13)2 = 169.
Ex.13. Which of the following numbers is
divisible by 3 ?
(i) 541326 (ii) 5967013
Sol.
(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is
divisible by 3.
Hence, 541326 is divisible by 3.
(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is
not divisible by 3.
Hence, 5967013 is not divisible by 3.
Ex.14.What least value must be assigned to * so that the number
197*5462 is r 9 ?
Sol.
Let the missing digit be x.
Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).
For (34 + x) to be divisible by 9, x must be replaced by 2 .
Hence, the digit in place of * must be 2.
Ex. 15. Which of the following numbers is divisible by 4 ?
(i) 67920594
(ii) 618703572
Sol.
(i) The number formed by the last two digits in the given number is
94, which is not divisible by 4.
Hence, 67920594 is not divisible by 4.
(ii) The number formed by the last two digits in the given number is
72, which is divisible by 4.
Hence, 618703572 is divisible by 4.
Ex. 16. Which digits should come in place of * and $ if the number
62684*$ is divisible by both 8 and 5 ?
Sol.
Since the given number is divisible by 5, so 0 or 5 must come in
place of $. But, a number ending with 5 is never divisible by 8. So, 0 will
replace $.
Now, the number formed by the last three digits is 4*0, which
becomes divisible by 8, if * is replaced by 4.
Hence, digits in place of * and $ are 4 and 0 respectively.
Ex. 17. Show that 4832718 is divisible by 11.
Sol. (Sum of digits at odd
places) - (Sum of digits at even places)
= (8
+ 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
Hence, 4832718 is
divisible by 11.
Ex. 18. Is 52563744 divisible by 24 ?
Sol.
24 = 3 x 8, where 3 and 8 are co-primes.
The sum of the digits
in the given number is 36, which is divisible by 3. So, the given number is divisible
by 3.
The number formed by
the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is
divisible by 8.
Thus, the given
number is divisible by both 3 and 8, where 3 and 8 are co-primes.
So, it is divisible by 3 x 8, i.e., 24.
Ex. 19. What least number must be added to 3000 to obtain a number
exactly divisible by 19 ?
Sol. On dividing 3000 by 19, we get 17
as remainder.
\Number to be added = (19 - 17) = 2.
Ex. 20. What least number must be subtracted from 2000 to get a
number exactly divisible by 17 ?
Sol. On dividing 2000 by 17, we get 11
as remainder.
\Required number to be subtracted = 11.
Ex. 21. Find the number which is nearest
to 3105 and is exactly divisible by 21.
Sol. On dividing 3105 by 21, we get 18
as remainder.
\Number to be added to 3105 = (21 - 18) = 3.
Hence, required number
= 3105 + 3 = 3108.
Ex. 22. Find the smallest number of 6 digits which is exactly
divisible by 111.
Sol. Smallest number of 6 digits is
100000.
On dividing 100000 by
111, we get 100 as remainder.
\Number to be added = (111 - 100) = 11.
Hence, required number
= 100011.
Ex. 23. On dividing 15968 by a certain
number, the quotient is 89 and the remainder is 37. Find the divisor.
Dividend - Remainder 15968-37
Sol.
Divisor = -------------------------- = ------------- = 179.
.Quotient 89
Ex. 24. A number when divided by 342 gives a remainder 47. When the
same number is divided by 19, what would be the remainder ?
Sol. On dividing the given number by 342, let k be the quotient and 47 as
remainder.
Then, number – 342k
+ 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
\The given number when divided by 19, gives (18k + 2) as quotient and
9 as remainder.
Ex. 25. A number being successively divided by 3, 5 and 8 leaves
remainders 1, 4
and 7 respectively. Find the respective remainders if the order of
divisors be reversed,
Sol.
|
3
|
X
|
|
|
5
|
y
|
- 1
|
|
8
|
z
|
- 4
|
|
|
1
|
- 7
|
\z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1)
= (3 x 79 + 1) = 238.
Now,
|
8
|
238
|
|
|
5
|
29
|
- 6
|
|
3
|
5
|
- 4
|
|
|
1
|
- 9,
|
\Respective remainders are 6, 4, 2.
Ex. 26. Find the remainder when 231
is divided by 5.
Sol. 210 = 1024. Unit digit of 210
x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].
\Unit digit of 231 is 8.
Now, 8 when
divided by 5, gives 3 as remainder.
Hence, 231 when
divided by 5, gives 3 as remainder.
Ex. 27. How many numbers between 11 and 90 are divisible by 7 ?
Sol. The required numbers are 14, 21, 28, 35, ....
77, 84.
This is an A.P. with a
= 14 and d = (21 - 14) = 7.
Let it contain n
terms.
Then, Tn =
84 => a + (n - 1) d = 84
=> 14 + (n - 1) x 7 = 84 or n = 11.
\Required number of terms = 11.
Ex. 28. Find the sum of all odd numbers
upto 100.
Sol. The given numbers are 1, 3, 5, 7, ..., 99.
This is an A.P. with a
= 1 and d = 2.
Let it contain n
terms. Then,
1 + (n - 1) x 2 = 99
or n = 50.
\Required sum = n (first term + last term)
2
= 50 (1 + 99) = 2500.
2
Ex. 29. Find the sum of all 2 digit
numbers divisible by 3.
Sol. All 2 digit numbers divisible by 3 are :
12, 51, 18, 21, ...,
99.
This is an A.P. with a
= 12 and d = 3.
Let it contain n
terms. Then,
12 + (n - 1) x 3 = 99
or n = 30.
\Required sum = 30 x (12+99) = 1665.
2
Ex.30.How many terms are there in 2,4,8,16……1024?
Sol.Clearly 2,4,8,16……..1024 form a GP.
With a=2 and r = 4/2 =2.
Let the number of terms
be n . Then
2 x 2n-1
=1024 or 2n-1 =512 = 29.
\n-1=9 or n=10.
Ex. 31. 2 + 22 + 23 + ... + 28 = ?
Sol. Given series is a G.P.
with a = 2, r = 2 and n = 8.
\sum = a(rn-1)
= 2 x (28 –1) = (2 x 255) =510
(r-1)
(2-1)
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