PREVIOUS NEXT
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 unjt 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.
PREVIOUS NEXT
No comments:
Post a Comment