1. The remainder when 1010 + 10100 + 101000 + . . . . . . + 101000000000 is divided by 7 is
a) 0
b) 1
c) 2
d) 5
Explanation: Number of terms in the series = 10.
(We can get it easily by pointing the number of zeros in power of terms.
In 1st term number of zero is 1, 2nd term 2, and 3rd term 3 and so on)
$$\frac{{{{10}^{10}}}}{7},$$ Written as, $$\frac{{{{\left( {7 + 3} \right)}^{\left( {4 \times 2 + 2} \right)}}}}{7}$$
The remainder will depend on $$\frac{{{3^2}}}{7}$$
So, remainder will be 2
$$\eqalign{ & \frac{{{{10}^{1000}}}}{7},\,{\text{remainder}} = 2 \cr & \frac{{{{10}^{10000}}}}{7},\,{\text{remainder}} = 1 \cr} $$
We get alternate 2 and 1 as remainder, five times each.
Required remainder is given by
$$\eqalign{ & \frac{{\left( {2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1} \right)}}{7} \cr & = \frac{{15}}{7} \cr} $$
Remainder when 15 is divided by 7 = 1
2. There are 154 beads in a rosary and all are coloured, either RED or BLUE or GREEN. The number of blue ones is three less than red and five more than green. The number of Red beads are:
a) 55
b) 47
c) 45
d) 52
Explanation: Let the number of red beads = x
Blue beads = (x - 3)
Green beads = (x - 8)
x + x - 3 + x - 8 = 154
3x - 11 = 154
x = 55
3. Recently, a small village, in Tamilnadu where only male shepherd reside with four sheep each, was devastated by Tsunami waves. Therefore 8 persons and 47 sheep were found to be dead and the person who luckily survived, left the village with one sheep each since 21 sheep were too injured to move so have left on their on luck, in the village. The number of sheep which were earlier in the village was :
a) 84 sheep
b) 120 sheep
c) 80 sheep
d) 90 sheep
Explanation: Let there be x people in the village so, then 4x sheep must be there.
No, number of person leaving the village,
= (x - 8) [as 8 person died in Tsunami]
And the number of sheep to be carried with the survived people,
4x - (47 + 21) = x - 8. [Here we are equating no. of alive sheep and no. of alive person. As 1 person has 4 sheep means 1 person = 4 sheep.]
3x = 60
x = 20
4x = 80
So, 80 sheep
4. A boy appeared in CAT for four consecutive year years, but coincidentally each time his net score was 75. He told me that there was $$\frac{1}{3}$$rd negative marking for every wrong answer and 1 mark was allotted for every correct answer. He has attempted all the questions every year, but certainly some answers have been wrong due conceptual problem. Which is not the total number of questions asked for CAT in any year, in that period?
a) 231
b) 163
c) 150
d) 123
Explanation: Since 75 is an integer, he must have to do (75 + 4x) problems, where x is a whole number. Now we put the value of x (0, 1, 2, 3, 4, 5 .........) to get the correct answer.
As he gets the net marks in integer it means when he does 3n question wrong, he losses actually 4n marks.
5. The highest power of 17 which can divide exactly the following:
(182 - 1) (184 - 1) (186 - 1) (188 - 1) . . . . (1816 - 1) (1818 - 1) is :
a) 1
b) 17
c) 9
d) 13
Explanation: Total number of terms in expression = 9
(182 - 1) = (18 - 1) × (18 + 1)
(184 - 1) = (182 + 1) × (182 - 1) = (182 + 1) × (18 - 1) × (18 + 1)
(186 - 1) = [(183)2 -1] = (183 + 1) (183 - 1) = 17 × k and so on.
There will be 9 times 17 in the whole expression as each term of expression gives one 17. Therefore for maximum power of 17 will be 9.
6. The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave $$\frac{1}{2}$$ of the diamonds he had then and 2 more besides. He escaped with one diamond. How many did he steal originally?
a) 36
b) 25
c) 640
d) None of these
Explanation: At last thief is left with one diamond.
Hence, the number of diamonds before he gave some diamonds to the third watchman,
$$\eqalign{ & x - \left( { {\frac{x}{2}} + 2} \right) = 1 \cr & {\kern 1pt} \frac{{ {x - 4} }}{2} = 1 \cr & {\kern 1pt} x = 6 \cr} $$
Hence, he had 6 diamonds before he gave 5 to the third watchman.
Similarly number of diamonds before giving to second watchman,
$$\eqalign{ & \frac{{ {x - 4} }}{2} = 6 \cr & {\kern 1pt} x = 16 \cr} $$
And number of diamonds before giving to the first watchman,
$$\eqalign{ & \frac{{ {x - 4} }}{2} = 16 \cr & {\kern 1pt} x = 36 \cr} $$
The thief has stolen 36 diamonds originally
7. Which one of the following is not a prime number?
a) 31
b) 61
c) 71
d) 91
Explanation: 91 is divisible by 7. So, it is not a prime number.
8. The remainder , when (22225555 + 55552222) is divided by 7, is
a) 4
b) 5
c) 0
d) 2
Explanation: an + bn is always divisible by (a + b) when n is odd.
(22225555 + 55552222) is always divisible by (2222 + 5555) = 7777
And 7777 is multiple of 7,
(22225555 + 55552222) is divisible by 7
9. A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9 and then back to the index finger for 10, middle finger for 11 and so on. She counted up 1994. She ended on her
a) Thumb
b) Index finger
c) Middle finger
d) Ring finger
Explanation: If the girl counts the way as given in the question,
The counting serial for thumb will be 1, 9, 17, 25 . . . . . .
Number 1992 ( as 1992 is divisible by 8, common difference of series formed by counting) will also fall on thumb.
Hence, Number 1994 will end on her middle finger.
10. How many pairs of natural numbers is there the difference of whose squares are 45.
a) 1
b) 2
c) 3
d) 4
Explanation: (x2 - y2) = 45
(x - y)(x + y) = 45
The factors of 45 possibles are 15, 3, 9, 5, 1 and 45
Hence, numbers are 9 and 6 or 7 and 2 or 23 and 22
Number of such pairs = 3